Statics

Mathematics Statics Exam night revision 2020-2021 1 The moment of a force about a point in 2D coordinate system ⃑ about a point O is denoted by 𝐌𝐨 ⃑⃑⃑⃑⃑⃑ The moment of a force 𝑭 ⃑⃑⃑⃑⃑⃑ ⃑ × ⃑𝑭 𝐌𝐨 = ⃑⃑⃑⃑⃑⃑ 𝑶𝑨 × ⃑𝑭 = 𝒓 ⃑ is the position vector of any point on the line of action of the force where 𝒓 ⃑⃑⃑⃑⃑⃑ × ⃑𝑭‖ = ‖𝑶𝑨 ⃑⃑⃑⃑⃑⃑ ‖. ‖𝑭 ⃑⃑⃑⃑⃑⃑ ‖ = ‖𝑶𝑨 ⃑ ‖ 𝒔𝒊𝒏𝜽 = ‖𝒓 ⃑ × ⃑𝑭‖ ‖𝐌𝐨 When the force acts such that the rotation in a clockwise direction then the moment is negative when rotates in anticlockwise direction then the moment is positive ⃑⃑⃑⃑⃑⃑𝟎 ‖ ‖𝑴 ⃑‖ ‖𝑭 ⃑⃑⃑⃑⃑⃑𝑨 = ⃑𝟎 𝒕𝒉𝒆𝒏 𝒕𝒉𝒆 𝒇𝒐𝒓𝒄𝒆 𝒑𝒂𝒔𝒔𝒆𝒔 𝒃𝒚 𝑨 𝑰𝒇 𝑴 𝑳𝑶 = 𝑰𝒇 ⃑⃑⃑⃑⃑⃑ 𝑴𝑨 = ⃑⃑⃑⃑⃑⃑ 𝑴𝑩 𝒕𝒉𝒆𝒏 𝒕𝒉𝒆 𝒇𝒐𝒓𝒄𝒆 𝒊𝒔 𝒑𝒂𝒓𝒂𝒍𝒍𝒆𝒍 𝒕𝒐 ⃡⃑⃑⃑⃑ 𝑨𝑩 𝑰𝒇 ⃑⃑⃑⃑⃑⃑ 𝑴𝑨 = − ⃑⃑⃑⃑⃑⃑ 𝑴𝑩 𝒕𝒉𝒆𝒏 𝒕𝒉𝒆 𝒇𝒐𝒓𝒄𝒆 𝒑𝒂𝒔𝒔𝒆𝒔 𝒃𝒚 𝒕𝒉𝒆 𝒎𝒊𝒅. 𝒑𝒐𝒊𝒏𝒕 𝒐𝒇 ̅̅̅̅ 𝑨𝑩 The general Theorem of moments :Algebraic Sum of moments of forces around a point equals the moment of their resultant about the same point. Principle of moments [Varignon’s theorem] The moment of a force about a point equals the sum of the moments of its components about the same point Let ⃑𝑭 = (𝑭𝒙 , 𝑭𝒚 , 𝑭𝒛 ) 𝒂𝒄𝒕𝒔 𝒂𝒕 𝑨(𝒂𝒙 , 𝒂𝒚 , 𝒂𝒛 ) 𝒊𝒕𝒔 𝒎𝒐𝒎𝒆𝒏𝒕 𝒂𝒃𝒐𝒖𝒕 𝑶(𝟎, 𝟎, 𝟎) At first get ⃑⃑⃑⃑⃑⃑ 𝑶𝑨 = (𝒂𝒙 , 𝒂𝒚 , 𝒂𝒛 ) ⃑⃑⃑⃑⃑⃑ × 𝑭 ⃑⃑⃑⃑⃑⃑𝒐 = 𝑶𝑨 ⃑ 𝑴 2 𝒊̂ ⃑⃑⃑⃑⃑⃑𝒐 = |𝒂𝒙 𝑴 𝑭𝒙 𝒋̂ 𝒂𝒚 𝑭𝒚 ̂ 𝒌 𝒂𝒛 | 𝑭𝒛 ̂ ⃑⃑⃑⃑⃑⃑𝒐 = [𝒚𝑭𝒛 − 𝒛𝑭𝒚 ]𝒊̂ + [𝒛𝑭𝒙 − 𝒙𝑭𝒛 ]𝒋̂ + [𝒙𝑭𝒚 − 𝒚𝑭𝒙 ]𝒌 𝑴 ̂ = [𝒎𝒐𝒎𝒆𝒏𝒕 𝒂𝒃𝒐𝒖𝒕 𝒙 − 𝒂𝒙𝒊𝒔]𝒊̂ + [𝒎𝒐𝒎𝒆𝒏𝒕 𝒂𝒃𝒐𝒖𝒕 𝒚 − 𝒂𝒙𝒊𝒔]𝒋̂ + [𝒎𝒐𝒎𝒆𝒏𝒕 𝒂𝒃𝒐𝒖𝒕 𝒛 − 𝒂𝒙𝒊𝒔]𝒌 The moment about an axis is zero if the line of action of the force cuts the axis or parallel to it 60 90 60 30 2𝐿 ξ3𝐿 90 120 60 30 30 ξ3 𝐿 2 30 ξ3𝐿 30 30 ξ3 𝐿 2 120 60 3 Parallel forces If we have two forces 𝑭𝟏 and 𝑭𝟐 act at A and B ,having the same direction then A B C ⃑⃑⃑ 𝐹1 ⃑⃑⃑⃑⃑ 𝑅 The magnitude of the resultant: 𝑹 = 𝑭𝟏 + 𝑭𝟐 ⃑⃑⃑ 𝐹2 The direction of the resultant : in their direction The point of action of the resultant : point C such that 𝑭𝟏 𝑭𝟐 = 𝑩𝑪 𝑨𝑪 Equilibrium of coplanar parallel forces Any system of coplanar parallel forces is in equilibrium State if 1. Resultant = 0. 2. Algebraic Sum of moments about any point = 0. If the two forces act in opposite directions and 𝑭𝟏 > 𝑭𝟐 𝑭𝟏 𝒂𝒄𝒕𝒔 𝒂𝒕 𝑨 𝒂𝒏𝒅 𝑭𝟐 𝒂𝒄𝒕𝒔 𝒂𝒕 𝑩 ⃑⃑⃑ 𝐹2 A B C 𝑅⃑ 𝐶 ⃑⃑⃑ 𝐹1 The magnitude of the resultant: 𝑹 = |𝑭𝟏 − 𝑭𝟐 | The direction of the resultant : in the direction of the greater force The point of action of the resultant: ̅̅̅̅ externally divides 𝑨𝑩 𝑭𝟏 𝑭𝟐 = 𝑩𝑪 𝑨𝑪 ̅̅̅̅,but on 𝑨𝑩 ⃡⃑⃑⃑⃑ , beside the greater force doesn’t lie on 𝑨𝑩 The force of friction The coefficient of friction gives the scale of roughness between any two bodies 4 smooth surface when it has no force of friction [its hypothetical it has no real existence] the coefficient of friction equals zero rough surface which has a friction force appears when the body is tends to move the coefficient of friction is more than zero The static friction force It's a hidden force appears as trying to make a body move on a rough plane The force of limiting static friction It's the friction force when the magnitude of the friction force is at its limiting value at which the body is a bout to move and denoted by Fs The coefficient of (limiting ) static friction [ 𝝁𝒔 ] the ratio between the limiting static friction [ Fs ] and the normal reaction(N) 𝝁𝒔 = 𝑭𝒔 𝑵 , 𝑭𝒔 = 𝝁𝒔 𝑵 It depends on the types of the body and the plane not the shape or the mass 𝟎 ≤ 𝑭𝒓 ≤ 𝝁𝒔 𝑵 If on rough plane and a force acts on it then 𝟎 < 𝑭𝒓 ≤ 𝝁𝒔 𝑵 The force of friction is zero on smooth planes or no force on a rough plane The maximum[limiting] force of friction when it about to move 5 When the body is under the action of a force less than the limiting force of friction then its at rest The kinetic force of friction If a body is moving on a rough surface then the force of friction is kinetic Fk 𝑭𝒌 = 𝝁𝒌 𝑵 The coefficient of the kinetic friction 𝝁𝒌 𝝁𝒌 = 𝑭𝒌 𝑵 the kinetic force of friction Fk is less than the static force of friction Fs 𝝁𝒌 < 𝝁𝒔 The resultant reaction Is The resultant of the force of friction and the normal reaction we call it the total reaction 𝑹\ = √𝑵𝟐 + (𝑭𝒓)𝟐 When the body is about to move 𝑭𝒓 = 𝑭𝒔 = 𝝁𝒔 𝑵 𝑹\ = √𝑵𝟐 + 𝑭𝒔 𝟐 = √𝑵𝟐 + (𝝁𝒔 𝑵)𝟐 = 𝑵√𝟏 + 𝝁𝒔 𝟐 The angle of friction 𝝀 is the angle between the normal reaction and the resultant reaction when the friction is limiting 𝑻𝒂𝒏 𝝀 = 𝑭𝒔 𝑵 6 𝒃𝒖𝒕 𝝁𝒔 = 𝑭𝒔 𝑵 𝒔𝒐 𝝁𝒔 = 𝑻𝒂𝒏 𝝀 𝒂𝒍𝒔𝒐 𝑹\ = 𝑵√𝟏 + 𝝁𝒔 𝟐 = 𝑵√𝟏 + (𝑻𝒂𝒏 𝝀)𝟐 = 𝑵√𝒔𝒆𝒄𝟐 𝝀 = 𝑵𝒔𝒆𝒄𝝀 we may use lami’s rule 𝑷 𝑾 𝑹\ = = 𝒔𝒊𝒏[𝟏𝟖𝟎 − 𝝀] 𝒔𝒊𝒏[𝟗𝟎 + 𝝀] 𝒔𝒊𝒏[𝟗𝟎] Properties of the static force of friction 1-It acts in a direction opposite to the motion 2-It equals to the force trying to make the body move and It increases as this force increases. 3- The force of friction is maximum when the body is about to move and we call it the limiting force of friction [ Fs ] 4-The body is moving when the force of action P is greater than the force of friction 5-The body is about to move when the force of action equals the force of static friction. 6- the ratio between the final force of friction [ Fs ] and the normal reaction constant and depends on the types of the two bodies not the shape or the weight and that ratio called the final coefficient of friction or[coefficient of friction] 𝝁𝒔 𝑹𝒆𝒎𝒆𝒎𝒃𝒆𝒓 𝒕𝒉𝒂𝒕 𝒕𝒉𝒆 𝒓𝒆𝒔𝒖𝒍𝒕𝒂𝒏𝒕 𝒐𝒇 𝒕𝒘𝒐 𝒇𝒐𝒓𝒄𝒆𝒔 𝒎𝒆𝒆𝒕𝒊𝒏𝒈 𝒂𝒕 𝒂 𝒑𝒐𝒊𝒏𝒕 𝑹 = √𝑭𝟐 𝟏 + 𝑭𝟐 𝟐 + 𝟐𝑭𝟏 𝑭𝟐 𝒄𝒐𝒔𝜶 , 𝑻𝒂𝒏 𝝑 = 𝑭𝟐 𝒔𝒊𝒏 𝜶 𝑭𝟏 + 𝑭𝟐 𝒄𝒐𝒔 𝜶 𝒊𝒇 𝒕𝒉𝒆 𝒕𝒘𝒐 𝒇𝒐𝒓𝒄𝒆𝒔 𝒂𝒓𝒆 𝒑𝒆𝒓𝒑𝒆𝒏𝒅𝒊𝒄𝒖𝒍𝒂𝒓 𝒕𝒉𝒆𝒏 𝜶 = 𝟗𝟎𝟎 𝒕𝒉𝒆𝒓𝒇𝒐𝒓𝒆 𝑹 = √𝑭𝟐 𝟏 + 𝑭𝟐 𝟐 , 𝑻𝒂𝒏 𝝑 = 𝑭𝟐 𝑭𝟏 7 Inclined rough plane If it body is about to slide down an inclined plane by the action of its weight only so 𝜶=𝝀 𝒂𝒏𝒅 𝝁𝒔 = 𝒕𝒂𝒏 𝜶 such that 𝜶 is the angle of inclination of the plane Relation between a and l a<l The body is at rest a=l The body is about to slide down the plane a>l The body is sliding down the plane The force which supports the body(the min. force to keep it from falling) If 𝜶 > 𝝀 So the body is moving down So The least force which prevents the body from sliding (its about to slide down) is called the force which supports the body 8 The couples The couple two forces Equal in magnitude, opposite in direction, and their lines of action are not on the same straight line The moment of a couple = one of the forces X the perpendicular distance between M=F.L The moment of a couple is a constant vector independent from the position of the point we take the moment about The moment of a couple acts in a direction perpendicular to the plane Equivalent couples Two couples are equivalent if 𝑴𝟏 = 𝑴𝟐 Equilibrium couples: Two couples are in equilibrium if 𝑴𝟏 + 𝑴𝟐 = 𝟎 If three forces act on a body are completely represented by the sides of a triangle in the same cyclic order then the system is equivalent to a couple The magnitude of the moment equals twice area multiplied by the scale 𝑴𝒂𝒏𝒈𝒏𝒊𝒕𝒖𝒅𝒆 𝒐𝒇 𝒎𝒐𝒎𝒆𝒏𝒕 = 𝟐[𝒂𝒓𝒆𝒂][𝒔𝒄𝒂𝒍𝒆] , 𝒔𝒄𝒂𝒍𝒆 = Equilibrium 𝑹=𝟎 𝒂𝒏𝒅 𝑴𝒂𝒏𝒚 𝒑𝒐𝒊𝒏𝒕 = 𝟎 𝒐𝒓 𝑴𝑨 = 𝑴𝑩 = 𝑴𝑪 = 𝟎 Couple 𝑹=𝟎 𝒂𝒏𝒅 𝑴𝒂𝒏𝒚 𝒑𝒐𝒊𝒏𝒕 ≠ 𝟎 oR 𝑴𝑨 = 𝑴𝑩 = 𝑴𝑪 ≠ 𝟎 𝑭𝒐𝒓𝒄𝒆 𝒍𝒆𝒏𝒈𝒕𝒉 9 The center of gravity 𝒙𝑮 = 𝒙𝟏 𝒘𝟏 + 𝒙𝟐 𝒘𝟐 + 𝒙𝟑 𝒘𝟑 + ⋯ . . +𝒙𝒏 𝒘𝒏 𝒘𝟏 + 𝒘𝟐 + 𝒘𝟑 + ⋯ … +𝒘𝒏 𝒚𝑮 = 𝒚𝟏 𝒘𝟏 + 𝒚𝟐 𝒘𝟐 + 𝒚𝟑 𝒘𝟑 + ⋯ . . +𝒚𝒏 𝒘𝒏 𝒘𝟏 + 𝒘𝟐 + 𝒘𝟑 + ⋯ … +𝒘𝒏 The center of gravity of the rigid body suspended freely lies on the vertical line passing through the suspension point to determine the center of gravity of a body we suspend it from a point and draw a vertical line then suspend it from a different point and draw another vertical line, then the intersection point of the two verticals is the point of action of the weight. The center of gravity of a fine symmetric lamina of uniform density lies on its axis of symmetry The center of gravity of a symmetric solid of uniform density lies on its plane of symmetry if existed The center of gravity of a uniform rod is its midpoint The mid. Point ( 𝒙𝟏 +𝒙𝟐 𝒚𝟏 +𝒚𝟐 𝟐 , 𝟐 ) The center of gravity of a uniform lamina in the form of a parallelogram [square ,rectangle, rhombus ] is the point of intersection of diagonals The center of gravity of a uniform lamina in the form of a circle is the center The center of gravity of a uniform lamina in the form of a triangle is the point of intersection of medians 10 ( 𝒙𝟏 + 𝒙𝟐 + 𝒙𝟑 𝒚𝟏 + 𝒚𝟐 + 𝒚𝟑 , ) 𝟑 𝟑 The center of gravity of a spherical crust(cortex) of a uniform density lies at the center The center of gravity of a uniform sphere lies at the center The center of gravity of a uniform cuboid[parallelepiped] lies at the geometric center[the point of intersection of diagonals] The center of gravity of a right circular cylindrical crust of a uniform density lies at the midpoint of its axis of symmetry The center of gravity of a right circular cylindrical of uniform density lies at the midpoint of its axis of symmetry The center of gravity of a uniform right prism lies at the midpoint of the axis parallel to its lateral edge and passing through the two centers of gravity of its two bases considering them as two fine laminas of uniform density The center of gravity of a uniform wire in the form of a semicircle is 𝟐𝒓 𝝅 The center of gravity of a uniform lamina in the form of a semicircle is The center of gravity of a uniform hemisphere 𝟑𝒓 𝟖 𝒂𝒃𝒐𝒗𝒆 𝒕𝒉𝒆 𝒄𝒆𝒏𝒕𝒆𝒓 above center 𝟒𝒓 𝟑𝝅 above center 11 The moment ̂ acts at 𝐀(−𝟑, 𝟏, 𝟐) find its moment vector about 𝐁(𝟐, 𝟐, −𝟏) 1]A force 𝐅 = 𝟐𝐢̂ − 𝟏𝐣̂ + 𝟑𝐤 and the length of perpendicular from B to the line of action of the force ⃑⃑⃑⃑⃑⃑ = 𝑨 ⃑⃑ − 𝑩 ⃑⃑ = (−𝟑, 𝟏, 𝟐) − (𝟐, 𝟐, −𝟏) = (−𝟓, −𝟏, 𝟑) 𝑩𝑨 𝒊̂ 𝒋̂ ⃑⃑⃑⃑⃑⃑ ⃑⃑⃑⃑⃑⃑ ⃑ 𝑴𝑩 = 𝑩𝑨 × 𝑭 = |−𝟓 −𝟏 𝟐 −𝟏 𝑳𝑩 = ̂ 𝒌 ̂ 𝟑| = (𝟎, 𝟐𝟏, 𝟕) = 𝟐𝟏𝒋̂ + 𝟕𝒌 𝟑 ⃑⃑⃑ 𝑩 ‖ √(𝟎)𝟐 + (𝟐𝟏)𝟐 + (𝟕)𝟐 ‖𝑴 = = ξ𝟑𝟓 𝒖𝒏𝒊𝒕 𝒐𝒇 𝒍𝒆𝒏𝒈𝒕𝒉 ⃑‖ ‖𝑭 √(𝟐)𝟐 + (−𝟏)𝟐 + (𝟑)𝟐 2]A force ⃑𝑭 = 𝟒𝒊̂ + 𝟑𝒋̂ acts at A(3,4). Find the moments vector of this force about B (-1,2) Find the length of perpendicular from B to the line of action of the force ⃑⃑⃑⃑⃑⃑ ⃑ − ⃑𝑩 ⃑ = (𝟑, 𝟒) − (−𝟏, 𝟐) = (𝟒, 𝟐) 𝑩𝑨 = ⃑𝑨 ⃑⃑⃑⃑⃑⃑ ̂ = 𝟒𝑲 ̂ 𝐌𝑩 = ⃑⃑⃑⃑⃑⃑ 𝑩𝑨 × ⃑𝑭 = (𝟒, 𝟐) × (𝟒, 𝟑) = [𝟏𝟐 − 𝟖]𝑲 ̂ The moments vector = 𝟒𝑲 The algebraic measure of the moments vector = 4 units ⃑⃑⃑⃑⃑⃑𝑩 ‖ = 𝟒 units The norm of the moments vector ‖𝐌 ̂ The body rotates in an anticlockwise direction with a moment in direction of 𝑲 𝑳𝑩 = ⃑⃑⃑⃑⃑⃑𝑩 ‖ ξ𝟎𝟐 + 𝟎𝟐 + 𝟒𝟐 𝟒 ‖𝑴 = = 𝒖𝒏𝒊𝒕𝒔 𝒐𝒇 𝒍𝒆𝒏𝒈𝒕𝒉 ⃑‖ ‖𝑭 ξ𝟒𝟐 + 𝟑𝟐 + 𝟎𝟐 𝟓 ̂ acts atA whose y coordinate is 2 , ⃑⃑⃑⃑⃑⃑ ̂ 3]A force 𝐅 = 𝟐𝐢̂ + 𝟑𝐣̂ − 𝐤 𝑴𝑶 = −𝟓𝒊̂ + 𝟑𝒋̂ − 𝒌 find A 𝐀(𝐱, 𝟐, 𝐳) 𝒊̂ ⃑⃑⃑⃑⃑⃑ × 𝑭 ⃑⃑⃑⃑⃑⃑ ⃑ = |𝒙 𝑴𝑶 = 𝑶𝑨 𝟐 ̂ 𝒋̂ 𝒌 𝟐 𝒁 | = (−𝟐 − 𝟑𝒛, 𝒙 + 𝟐𝒛 , 𝟑𝒙 − 𝟒) = (−𝟓, 𝟑, −𝟏) 𝟑 −𝟏 −𝟐 − 𝟑𝒛 = −𝟓 𝒔𝒐 𝒛 = 𝟏 𝟑𝒙 − 𝟒 = −𝟏 𝒔𝒐 𝒙 = 𝟏 𝒔𝒐 𝑨(𝟏, 𝟐, 𝟏) ̂ acts at A whose position vector from the origin 𝐫 = (𝟑, 𝟏, 𝟏) , the 4]A force 𝐅 = 𝐤𝐢̂ + 𝐦𝐣̂ − 𝟐𝐤 components of the moments of the force about x-axis is -1 and about y-axis is -8 find k and m 12 ̂ 𝒊̂ 𝒋̂ 𝒌 ⃑⃑⃑⃑⃑⃑𝒐 = ⃑⃑⃑⃑⃑ ⃑ × ⃑𝑭 = | 𝟑 𝟏 𝑴 𝒐𝑨 × ⃑𝑭 = 𝒓 𝟏 | = (−𝟐 − 𝒎 , 𝟔 + 𝒌 , 𝟑𝒎 − 𝒌) 𝑲 𝒎 −𝟐 component of moment about x-axis= - 1 -2-m=-1 so m=-1 component of moment about y-axis = - 8 6+k=-8 so k=-14 ⃑ 𝒂𝒄𝒕𝒔 𝒊𝒏 𝒕𝒉𝒆 𝒑𝒍𝒂𝒏𝒆 𝒔𝒖𝒄𝒉 𝒕𝒉𝒂𝒕 𝑴𝑨(𝟏,𝟐) = 𝟎 , 𝑴𝑩(𝟑,𝟒) = 𝑴𝒄(𝟓,𝟕) = 𝟏𝟎 𝟓] 𝑨 𝒇𝒐𝒓𝒄𝒆 𝑭 𝒇𝒊𝒏𝒅 𝒕𝒉𝒆 𝒇𝒐𝒓𝒄𝒆 𝒗𝒆𝒄𝒕𝒐𝒓 let ⃑𝑭 = 𝒍𝒊 + 𝒎𝒋 ⃑𝑭 𝒂𝒄𝒕𝒔 𝒂𝒕 𝑨(𝟏, 𝟐) ⃑ 𝒔𝒐 (– 𝟐, −𝟐) × (𝒍, 𝒎) = 𝟏𝟎𝒌 ⃑ 𝒔𝒐 − 𝟐𝒎 + 𝟐𝒍 = 𝟏𝟎 ⃑𝑴 ⃑⃑ 𝑩 = ⃑⃑⃑⃑⃑⃑ 𝑩𝑨 × ⃑𝑭 = 𝟏𝟎𝒌 ⃑ 𝒔𝒐 (– 𝟒, −𝟓) × (𝒍, 𝒎) = 𝟏𝟎𝒌 𝒔𝒐 − 𝟒𝒎 + 𝟓𝒍 = 𝟏𝟎 ⃑𝑴 ⃑⃑ 𝒄 = ⃑⃑⃑⃑⃑ 𝑪𝑨 × ⃑𝑭 = 𝟏𝟎𝒌 ⃑ = −𝟏𝟎𝒊 − 𝟏𝟓𝒋 𝑭 L= -10 and m= -15 ⃑⃑⃑𝟏 = 𝟐𝐢̂ + 𝟑𝐣̂, 𝐅 ⃑⃑⃑𝟐 = 𝟓𝐢̂, ⃑⃑⃑ 6] Forces 𝐅 𝐅𝟑 = −𝟔𝐣̂ act at A (2,1 ). Then Find the moment of their resultant about B(-1,4 ) ⃑⃑⃑⃑⃑⃑ ⃑ − ⃑𝑩 ⃑ = (𝟐, 𝟏) − (−𝟏, 𝟒) = (𝟑, −𝟑) 𝑩𝑨 = ⃑𝑨 ⃑⃑ = 𝟐𝟕 − 𝟑𝐣̂ 𝑹 ⃑⃑⃑⃑⃑⃑ ⃑⃑⃑⃑⃑⃑ × 𝑹 ⃑⃑ = (𝟑, −𝟑) × (𝟕, −𝟑) = 𝟏𝟐𝑲 ̂ 𝑴𝑩 = 𝑩𝑨 ⃑⃑⃑𝟏 = 𝟐𝐢̂ + 𝟑𝐣̂ acts at A(2,5) and𝐅 ⃑⃑⃑𝟐 = 𝐢̂ − 𝟓𝐣̂ acts at B ( 1,-3 ). Find the sum of moments 7] Two forces𝐅 of these forces about c ( -1,-2 ) and the length of ⊥ from C to the line of action of their resultant. ⃑⃑⃑⃑⃑ ⃑ − ⃑𝑪 = (𝟑, 𝟕) , ⃑⃑⃑⃑⃑⃑ ⃑ − ⃑𝑪 = (𝟐, −𝟏) 𝑪𝑨 = ⃑𝑨 𝑪𝑩 = ⃑𝑩 ⃑⃑⃑⃑⃑𝒄 = ⃑⃑⃑⃑⃑ ⃑⃑⃑⃑𝟏 + ⃑⃑⃑⃑⃑⃑ ⃑⃑⃑⃑𝟐 = (𝟑, 𝟕) × (𝟐, 𝟑) + (𝟐, −𝟏) × (𝟏, −𝟓) 𝑴 𝑪𝑨 × 𝑭 𝑪𝑩 × 𝑭 ̂ + [−𝟏𝟎 + 𝟏]𝑲 ̂ = −𝟏𝟒𝑲 ̂ = [𝟗 − 𝟏𝟒]𝑲 ⃑⃑ = 𝑭 ⃑⃑⃑⃑𝟏 + 𝑭 ⃑⃑⃑⃑𝟐 = (𝟑, −𝟐) 𝑹 𝑳𝑪 = ⃑⃑⃑⃑⃑𝒄 ‖ 𝟏𝟒 𝟏𝟒 ‖𝑴 = = ⃑⃑ ‖ ‖𝑹 √𝟑𝟐 + (−𝟐)𝟐 ξ𝟏𝟑 13 ⃑⃑⃑𝟐 = 𝟓𝐢̂ + 𝟐𝐣̂ , 𝐅 ⃑⃑⃑𝟑 = −𝟑𝐢̂ + 𝟐𝐣̂ act at A ( 1,1 ) 8] Forces ⃑⃑⃑ 𝐅𝟏 = 𝟐𝐢̂ − 𝐣̂ , 𝐅 Using moments show that the line of action of the resultant is parallel to the line joining B(2,4) and C(6,7) ⃑⃑ = 𝐅 ⃑⃑⃑𝟏 + 𝐅 ⃑⃑⃑𝟐 + 𝐅 ⃑⃑⃑𝟑 = (𝟒, 𝟑) 𝑹 ̂ ⃑⃑⃑⃑⃑⃑ ⃑⃑ = (−𝟏, −𝟑) × (𝟒, 𝟑) = 𝟗𝐤 𝑴𝑩 = ⃑⃑⃑⃑⃑⃑ 𝑩𝑨 × 𝑹 ̂ ⃑⃑⃑⃑⃑𝒄 = ⃑⃑⃑⃑⃑ ⃑⃑ = (−𝟓, −𝟔) × (𝟒, 𝟑) = 𝟗𝐤 𝑴 𝑪𝑨 × 𝑹 ⃑⃑⃑⃑⃑⃑ 𝑴𝑩 = ⃑⃑⃑⃑⃑⃑ 𝑴𝑪 𝒕𝒉𝒆 𝒇𝒐𝒓𝒄𝒂 𝒊𝒔 𝒑𝒂𝒓𝒂𝒍𝒍𝒆𝒍 𝒕𝒐 ⃡⃑⃑⃑⃑ 𝑩𝑪 𝟗] ⃑𝑭 = (𝟑, 𝟑)𝒂𝒄𝒕𝒔 𝒂𝒕 𝑨(𝟎, 𝟎)𝒔𝒉𝒐𝒘 𝒕𝒉𝒂𝒕 𝒕𝒉𝒆 𝒍𝒊𝒏𝒆 𝒐𝒇 𝒂𝒄𝒕𝒊𝒐𝒏 𝒐𝒇 𝒕𝒉𝒆 𝒇𝒐𝒓𝒄𝒆 𝒑𝒂𝒔𝒔𝒆𝒔 𝒃𝒚 𝒕𝒉𝒆 𝒎𝒊𝒅 𝒑𝒐𝒊𝒏𝒕 𝒐𝒇 ̅̅̅̅ 𝑪𝑫, 𝒄(𝟑, 𝟎), 𝑫(−𝟑, 𝟎) ⃑⃑⃑⃑⃑⃑ ⃑ = (−𝟑, 𝟎)𝑿(𝟑, 𝟑) = −𝟗𝑲 ⃑⃑⃑ 𝑴𝑪 = ⃑⃑⃑⃑⃑ 𝑪𝑨𝑿𝑭 ⃑⃑⃑⃑⃑⃑ ⃑ = (𝟑, 𝟎)𝑿(𝟑, 𝟑) = 𝟗𝑲 ⃑⃑⃑ 𝑴𝑫 = ⃑⃑⃑⃑⃑⃑ 𝑫𝑨𝑿𝑭 The force passes by the midpoint of ̅̅̅̅ 𝑪𝑫 ⃑⃑⃑𝟏 = 𝟑𝐢̂ + 𝟒𝐣̂ acts at A(1,2) , 𝐅 ⃑⃑⃑𝟐 = 𝟓𝐢̂ − 𝐣̂ act at B ( -1,3 ) 10] Forces 𝐅 Find the equation of the line of action of the resultant and point of intersection with y-axis ̂ ⃑⃑⃑⃑⃑⃑ ⃑⃑⃑⃑𝟏 + ⃑⃑⃑⃑⃑⃑ ⃑⃑⃑⃑𝟐 = (𝟏, 𝟐) × (𝟑, 𝟒) + (−𝟏, 𝟑) × (𝟓, −𝟏) = −𝟏𝟔𝒌 𝑴𝑶 = ⃑⃑⃑⃑⃑⃑ 𝑶𝑨 × 𝑭 𝑶𝑩 × 𝑭 ⃑⃑ = 𝟖𝐢̂ + 𝟑𝐣̂, let the resultant passes by C( x,y ) 𝐑 M of resultant about origin ⃑⃑⃑⃑⃑⃑ 𝑿𝑹 ⃑⃑⃑⃑⃑⃑𝟎 = 𝑶𝑪 ⃑⃑ = −𝟏𝟔𝑲 ⃑⃑⃑ 𝒔𝒐 (𝒙, 𝒚)𝑿(𝟖, 𝟑) = −𝟏𝟔𝑲 ⃑⃑⃑ 𝑴 𝟑𝒙 − 𝟖𝒚 = −𝟏𝟔 To cut y-axis then x=0 so y=2 point (0,2) 11] True or false a]the moment of a force about an axis will vanish if the force cuts the axis or parallel to it True or false True ⃑ × ⃑⃑⃑⃑⃑⃑ b] the moment of force F acts at A about B is ⃑⃑⃑⃑⃑⃑ 𝑴𝑩 = ⃑⃑⃑⃑⃑⃑ 𝑩𝑨 × ⃑𝑭 = −𝑭 𝑩𝑨 = ⃑𝑭 × ⃑⃑⃑⃑⃑⃑ 𝑨𝑩 True 14 11] A force of magnitude 130Newton ̅̅̅̅ acts along𝑨𝑩 Find its moment about D A(0,0,3) , B(4,12,0) ,D(0,12,0) ⃑⃑⃑⃑⃑⃑ = 𝑩 ⃑⃑ − 𝑨 ⃑⃑ = (𝟒, 𝟏𝟐, 𝟎) − (𝟎, 𝟎, 𝟑) = (𝟒, 𝟏𝟐, −𝟑) 𝑨𝑩 ⃑⃑⃑⃑⃑⃑ 𝑨𝑩∗ = ⃑⃑⃑⃑⃑⃑ (𝟒, 𝟏𝟐, −𝟑) (𝟒, 𝟏𝟐, −𝟑) 𝑨𝑩 = = ⃑⃑⃑⃑⃑⃑ ‖ √(𝟒)𝟐 + (𝟏𝟐)𝟐 + (−𝟑)𝟐 𝟏𝟑 ‖𝑨𝑩 ⃑𝑭 = [𝟏𝟑𝟎]𝑨𝑩 ⃑⃑⃑⃑⃑⃑ ∗ = [𝟏𝟑𝟎] (𝟒, 𝟏𝟐, −𝟑) = (𝟒𝟎, 𝟏𝟐𝟎, −𝟑𝟎) 𝟏𝟑 ⃑⃑⃑⃑⃑⃑ ⃑ − ⃑𝑫 ⃑ = (𝟒, 𝟏𝟐, 𝟎) − (𝟎, 𝟏𝟐, 𝟎) = (𝟒, 𝟎, 𝟎) 𝑫𝑩 = ⃑𝑩 ̂ 𝒊̂ 𝒋̂ 𝒌 ̂ ⃑⃑⃑⃑⃑⃑ ⃑⃑⃑⃑⃑⃑ ⃑ 𝑴𝑫 = 𝑫𝑩 × 𝑭 = | 𝟒 𝟎 𝟎 | = (𝟎 , 𝟏𝟐𝟎 , 𝟒𝟖𝟎) = 𝟏𝟐𝟎𝒋̂ + 𝟒𝟖𝟎𝒌 𝟒𝟎 𝟏𝟐𝟎 −𝟑𝟎 12] Calculate the moment of the force 52 acts along BC about A 𝐴(0,0,0) 𝐵(0,12,0) 𝑐(−4,0,3) ‖𝐹 ‖ = 52 ⃑ = (−4, −12,3) ∴ ⃑⃑⃑⃑⃑ 𝐵𝐶 = 𝐶 − 𝐵 (−4, −12,3) 𝐹̅ = ‖𝐹 ‖ . 𝐵𝐶 ∗ = 52 = (−16, −48 , 12 ) 13 ⃑⃑⃑⃑⃑ 𝐵𝐶 𝐵𝐶 ∗ = ⃑⃑⃑⃑⃑ || ||𝐵𝐶 𝑖̂ 𝑗̂ 𝑘̂ ⃑⃑⃑⃑⃑ ⃑⃑⃑⃑⃑ 𝑀𝐴 = 𝐴𝐵 × 𝐹 = | 0 12 0| −16 −48 12 = (𝑖̂)(144) + (−𝑗̂)(0) + (𝑘̂)(192) ̂ = 144𝑖̂ + 192𝐾 15 13] Calculate the moment of the force of magnitude 8 N acts at C about A 𝜃𝑥 = 60° 𝜃𝑦 = 90° 𝜃𝑧 = 150° 𝐹̅ = |𝐹| (cos 𝜃𝑥 , cos 𝜃𝑦 , cos 𝜃𝑧 ) 1 ξ3 = 8 ( , 0 , − ) = ( 4 , 0 , −4ξ3) 2 2 ̅̅̅̅ = (0,6,9) 𝐴 = (0,0, −6) 𝐶 (0,6,3) ∴ 𝐴𝐶 − 𝑖 𝑗 𝑘 ̅̅̅̅ ̅ 𝑀𝐴 = 𝐴𝐶 𝑥 𝐹 = |0 6 9 | 4 0 −4ξ3 = (−24ξ3 𝑖̂) + (−36)(−𝑗̂) − 24𝑘 = −24ξ3𝑖̂ + 36 𝑗̂ − 24𝑘̂ 14] Determine the sum of moments of the forces about O 𝐶 = (12,0,9) 𝐼 = (0,13,9) 𝑓̅1 = |𝐹1 |𝐶𝐷∗ = 49 𝐷(12,13,9) 𝐵(12,13,0) 𝐶𝐷(0,13,0) 𝐼𝐵 = (12,0, −9) (0,13,0) = (0, 49,0) ξ132 (12,0, −9) 𝑓2̅ = |𝐹2 |𝐼𝐵∗ = 30 = (24,0, −18) ξ122 + 92 𝑀𝑜 = ̅̅̅̅ 𝑂𝐶 𝑥𝑓1̅ + ̅̅̅ 𝑂𝐼 + 𝑓2̅ − − 𝑖 𝑗 𝑘 𝑖 𝑗 𝑘 = |12 0 9| + | 0 13 9 |= 0 49 0 24 0 −18 (−441 , 0, 588) + (−234,216, −312) = ( −675 , 216 , 276 ) 16 16] find the algebraic measure of the moment about o 𝑀𝑜 = (−11)(30 − 𝑍) 𝑦 = 28 cos 60 = 14 → 𝑍 = 𝑋 → 𝑥 = 25 − 𝑦 = 11 𝑐𝑚 (30 − 11) ∴ 𝑀𝑜 = (−11) = −2.09 𝑁. 𝑚 100 17] find the algebraic measure of the moment about o 𝑀𝑜 = (−38 sin 30)(6) +(−38 cos 30)(5ξ3) = −399 𝑁. 𝑐𝑚 30-z Z X Y 17 18] ABC is a right angled  at B , AB = 6 , BC = 8. force, ⃑𝑭 acts in its plane, MA =MB = 60 , MC = -60. Find f and its line of action .also find the force vector as the moment about A equals negative the moment about C then the force passes by the midpoint of ̅̅̅̅ 𝑨𝑪 which is Y as the moment about B equals negative the moment about C ̅̅̅̅ then the force passes by the midpoint of 𝑩𝑪 which is x as the moment about B positive then the force must act in an anti-clock wise direction relative to ̅̅̅̅ [line of action] B then the force acts along 𝑿𝒀 to get the magnitude 𝑴𝑩 = [𝑭][𝑩𝒙] = 𝟔𝟎 𝑴𝑩 = [𝑭][𝟒] = 𝟔𝟎 𝒕𝒉𝒆𝒏 𝑭 = 𝟏𝟓 𝑵𝒆𝒘𝒕𝒐𝒏 A(0,0),B(6,0),C(6,8) then x(6,4),Y(3,4) ⃑⃑⃑⃑⃑ 𝑿𝒀 = (𝟑, 𝟒) − (𝟔, 𝟒) = (−𝟑, 𝟎) ⃑ = 𝟏𝟓 𝑿𝒀 ⃑⃑⃑⃑⃑ ∗ = 𝟏𝟓. 𝑭 (−𝟑, 𝟎) ⃑⃑⃑⃑⃑ 𝒙𝒚 = 𝟏𝟓. = (−𝟏𝟓, 𝟎) ‖𝒙𝒚 ⃑⃑⃑⃑⃑ ‖ 𝟑 18 19] A force acts in the plane xy,its moment about O is 198N.cm, its moment about A is -170 N.cm , and its moment about B is zero . Find 𝐅 𝐌𝐨 = 𝟏𝟗𝟖 𝐍. 𝐦 𝐅 = (𝐥, 𝐦) 𝐁(𝟗, 𝟎) 𝐌𝐀 = 𝟏𝟕𝟎 𝐀(𝟎, 𝟖) ̂ ̅̅̅̅ ̅̅̅̅ ̅ 𝐌𝐁 = 𝟎 𝐌𝐀 = 𝐀𝐁 ∗ 𝐅 = (𝟗, −𝟖) ∗ (𝐥, 𝐦) = −𝟏𝟕𝟎 𝐤 𝟗𝐦 + 𝟖𝐥 = −𝟏𝟕𝟎 → 𝟏 ̂ 𝐌𝐨 = 𝐎𝐁 ∗ 𝐅 = (𝟗, 𝟎) ∗ (𝐥, 𝐦) = 𝟏𝟗𝟖𝐤 𝟗𝐦 = 𝟏𝟗𝟖 → 𝐦 = 𝟐𝟐 , 𝐢𝐧 𝟏 → ∴ 𝐥 = −𝟒𝟔 𝐅̅ = (−𝟒𝟔, 𝟐𝟐) = −𝟒𝟔 𝐢̂ + 𝟐𝟐 𝐣̂  20]ABCD is a rhombus whose side is 10 cm m ( A) = 60 o , forces of magnitudes 7, 16, 15, 11N. act along BA, BC , DC and DB respectively. Find the alg. Sum of moments about A. 𝑴𝑨 = [−𝟏𝟏][𝟓ξ𝟑] + [𝟏𝟔 𝐬𝐢𝐧 𝟔𝟎][𝟓] + [𝟏𝟔 𝐜𝐨𝐬 𝟔𝟎 ][𝟓ξ𝟑] +[−𝟏𝟓 𝐜𝐨𝐬 𝟔𝟎][𝟓ξ𝟑] + [−𝟏𝟓 𝐬𝐢𝐧 𝟔𝟎] [𝟓] = − 𝟓𝟎ξ𝟑 𝑵𝒆𝒘𝒕𝒐𝒏 . 𝒄m 19 21]ABCD is a trapezium ,m(B)=900, AD//BC ,AB=12cm,BC=8 cm,AD=9cm,forces of magnitudes F1,60,F2,30 gm.wt act along ⃑⃑⃑⃑⃑ 𝐁𝐀, ⃑⃑⃑⃑⃑ 𝐁𝐂, ⃑⃑⃑⃑⃑ 𝐃𝐂, ⃑⃑⃑⃑⃑⃑ 𝐃𝐀 resp. Find the values of F1,F2 given that MA=Mc=0. 8 60 B 𝑴𝑪 = [−𝟑𝟎][𝟏𝟐] + [ 𝑭𝟏 ][𝟖] = 𝟎 C 12 𝒔𝒐 𝑭𝟏 = 𝟒𝟓 𝒈𝒎. 𝒘𝒕 𝑴𝑨 = [−𝟔𝟎][𝟏𝟐] + [ 𝑭𝟐 𝐬𝐢𝐧 𝜽][𝟗] = 𝟎 [−𝟔𝟎][𝟏𝟐] + [ 𝑭𝟐 ( 𝒔𝒐 𝑭𝟐 = 𝟏𝟐 ξ𝟏𝟒𝟓 F2 F1 12 )] [𝟗] = 𝟎 𝟐𝟎ξ𝟏𝟒𝟓 𝒈𝒎. 𝒘𝒕 𝟑 30 A 8 ξ145 𝛉 1 D 22]ABCDEH is a regular hexagon whose side length 10 cm. Forces 3, 4, 5, 6, 7 and 8N. act along. AB , CB , DC , DE , EH and HA . Find the alg. Sum of moments of these forces about the center of the hexagon, and about the vertex A. The length of the perpendicular from the Point of intersection of diagonals to each side Equals ξ𝟑 [𝟏𝟎] 𝟐 = 𝟓ξ 𝟑 𝑴𝒙 = [𝟑][𝟓ξ𝟑] [−𝟒][𝟓ξ𝟑] +[−𝟓][𝟓ξ𝟑] +[𝟔][𝟓ξ𝟑] +[𝟕][𝟓ξ𝟑] +[𝟖][𝟓ξ𝟑] = 𝟕𝟓ξ𝟑 𝑵. 𝒄𝒎 𝑴𝑨 = [−𝟒][𝟓ξ𝟑] + [−𝟓][𝟏𝟎ξ𝟑] + [𝟔][𝟏𝟎ξ𝟑] + [𝟕][𝟓ξ𝟑] = 𝟐𝟓ξ𝟑 𝑵𝒆𝒘𝒕𝒐𝒏. 𝒄𝒎 20 Parallel forces 1)Two parallel forces having the same direction, the distance between their lines of action is 30cm. If the magnitudes of their resultant is 80 Newton , and its line of action is distant 5cm from the first force . Find the magnitudes of the two forces. A C let ⃑𝑪 is a unit vector in their direction 𝐹1 ⃑⃑ = 𝑭 ⃑⃑⃑⃑𝟏 + 𝑭 ⃑⃑⃑⃑𝟐 𝑹 ⃑⃑⃑ 𝑅 B 𝐹2 𝐶 𝟖𝟎 = 𝑭𝟏 + 𝑭𝟐 AC=5 cm so BC=25 cm 𝑭𝟏 𝑩𝑪 𝑭𝟏 𝟐𝟓 = 𝒔𝒐 = 𝒕𝒉𝒆𝒏 𝑭𝟏 = 𝟓𝑭𝟐 𝑭𝟐 𝑨𝑪 𝑭𝟐 𝟓 𝑭𝟏 + 𝑭𝟐 = 𝟖𝟎 𝟓𝑭𝟐 + 𝑭𝟐 = 𝟖𝟎 𝟔𝑭𝟐 = 𝟖𝟎 𝑭𝟐 = 𝟒𝟎 𝟐𝟎𝟎 𝑵𝒆𝒘𝒕𝒐𝒏 , 𝑭𝟏 = 𝑵𝒆𝒘𝒕𝒐𝒏 𝟑 𝟑 2)Two forces 10 Newton and 25Newton act at A,B such that AB=40 cm. if the two forces are in opposite direction then Find their resultant. 𝑅⃑ 25 let ⃑𝑪 is a unit vector in the direction of the greater force 𝐶 A B C 10 The magnitude of the resultant: 𝑹 = |𝑭𝟏 − 𝑭𝟐 | = 𝟏𝟓 𝑵𝒆𝒘𝒕𝒐𝒏 The direction of the resultant : in the direction of the greater force [ the second force] The point of action of the resultant: ̅̅̅̅,but on ⃡⃑⃑⃑⃑ doesn’t lie on 𝑨𝑩 𝑨𝑩, beside the greater force at C , let BC=L 21 𝑭𝟏 𝑭𝟐 = 𝑩𝑪 𝑨𝑪 so 𝟏𝟎 𝟐𝟓 = it acts at a point 𝑳 𝟒𝟎+𝑳 𝟖𝟎 𝟑 𝒕𝒉𝒆𝒏 𝟒𝟎𝟎 + 𝟏𝟎𝑳 = 𝟐𝟓 𝒕𝒉𝒆𝒏 𝟏𝟓𝑳 = 𝟒𝟎𝟎 𝒔𝒐 𝑳 = 𝟖𝟎 𝟑 𝒄𝒎 𝒄𝒎 from B (3)Two Parallel forces 20 Newton and F Newton act at A,B ,their resultant is 35 Newton, the distance between the line of action of the known force and the line of action of the resultant is 15 cm, Find F and the distance between the two forces if a- the resultant and the known force are in the same direction. b- the resultant and the known force are in opposite directions. a-The resultant and the given force are in the same direction let ⃑𝑪 is a unit vector in the direction of the given force ⃑⃑ = 𝑭 ⃑⃑⃑⃑𝟏 + 𝑭 ⃑⃑⃑⃑𝟐 𝑹 ⃑ = 𝟐𝟎𝒄 ⃑ + ⃑𝑭 𝟑𝟓𝒄 A 15 ⃑𝑭 = 𝟏𝟓𝒄 ⃑ C B 20 the two forces are having the same direction 𝑅 = 35 𝑭𝟏 𝑭𝟐 = 𝑩𝑪 𝑨𝑪 , 𝟐𝟎 𝟏𝟓 = 𝑪𝑩 𝟏𝟓 15 , 𝑪𝑩 = 𝟐𝟎𝒄𝒎 𝒔𝒐 𝑨𝑩 = 𝟑𝟓𝒄𝒎 𝐶 b-The resultant and the given force are in opposite directions let ⃑𝑪 is a unit vector in the direction of the given force F ⃑⃑ = 𝑭 ⃑⃑⃑⃑𝟏 + 𝑭 ⃑⃑⃑⃑𝟐 𝑹 ⃑ so 𝑭 ⃑ = −𝟓𝟓𝒄 ⃑ = 𝟐𝟎𝒄 ⃑ +𝑭 ⃑ −𝟑𝟓𝒄 A the two forces are in opposite directions 20 and the resultant opposite to the given force 𝑭𝟏 𝑭𝟐 = 𝑩𝑪 𝑨𝑪 , 𝟐𝟎 𝟓𝟓 = 𝑪𝑩 𝟏𝟓 , 𝑪𝑩 = 𝟔𝟎 𝟏𝟏 𝒄𝒎 𝒕𝒉𝒆𝒏 𝑨𝑩 = 𝟏𝟎𝟓 𝟏𝟏 𝒄𝒎 B 𝑅 = 55 C 15 𝐶 22 (4)Two parallel forces in the same direction of magnitudes 𝑭, 𝟐𝑭 act at A and B. If 2F moves parallel ⃑⃑⃑⃑⃑⃑ . to itself a distance X on𝑨𝑩 𝟐 Prove that the resultant moves a distance 𝑿 in the same direction. 𝟑 𝑭 𝟐𝑭 = 𝑪𝑩 so 𝑪𝑨 R 𝑪𝑨 = 𝟐𝑪𝑩 2f f A \ R \ A C [𝑨𝑪 + 𝑳] = 𝟐 [𝑩𝑪 − 𝑳 + 𝒙] 𝑨𝑪 + 𝑳 = 𝟐 𝑩𝑪 − 𝟐𝑳 + 𝟐𝒙 𝑳 = −𝟐𝑳 + 𝟐𝒙 𝟑𝑳 = 𝟐𝒙 𝑳= X L 𝟐 𝑿 𝟑 C\ B B 2f f 𝑭 𝑪𝑩 = \ 𝟐𝑭 𝑪𝑨 𝑪\ 𝑨 = 𝟐 𝑪\ 𝑩\ C B\ 23 5) A,B,C,D and E are five points on one St. line, where 2AB =2 BC =DE=4, CD = 3, Forces 10, 20, 5, 8 and 10 act at A,B,C,D and E. St. The last two forces act in direction opp. to the first 3 forces. Find the resultant.    R = (10 + 20 + 5 − 8 − 10)C = 17C R = 17 Let the resultant acts at X 10 10 20 R 5 ( AX )(17) = 0 + 2  20 + 4  5 − 8  7 −10 11 (17) AX = 40 + 20 − 56 −110 = −106 2 X − 106 AX = = −6.23 17 A 2 B 3 X D 4 C 8  X  AE   X  EA, X  EA E 10 AX = 6.3 ⃑⃑⃑𝟐 ‖ = 𝟓ξ𝟐 𝐢𝐧 𝐨𝐩𝐩𝐨𝐬𝐢𝐭𝐞 𝐝𝐢𝐫𝐞𝐜𝐭𝐢𝐨𝐧 𝐨𝐟 ⃑⃑⃑ (𝟔)𝐓𝐰𝐨 𝐟𝐨𝐫𝐜𝐞𝐬 ⃑⃑⃑ 𝐅𝟏 = 𝐢 + 𝟐 𝐣 𝒂𝒏𝒅 ‖𝐅 𝐅𝟏 𝒇𝒊𝒏𝒅 ⃑⃑⃑ 𝐅𝟐 𝑭𝟐 = −𝟓ξ𝟐 𝑭𝟏 ∗ = −𝟓ξ𝟐 [ 𝟏 ξ𝟓 𝒊+ 𝟐 ξ𝟓 𝒋]= −ξ𝟏𝟎 𝒊 − 𝟐ξ𝟏𝟎 𝒋 (𝟕)𝐓𝐰𝐨 𝐩𝐚𝐫𝐚𝐥𝐥𝐞𝐥 𝐟𝐨𝐫𝐜𝐞𝐬 ⃑⃑⃑ 𝐅𝟏 𝐚𝐧𝐝 ⃑⃑⃑ 𝐅𝟐 𝐚𝐜𝐭 𝐚𝐭 𝐀 𝐚𝐧𝐝 𝐁 , 𝐀𝐁 = 𝟐𝟎 𝐜𝐦 ⃑⃑⃑ 𝐅𝟏 = 𝟑𝐢 + 𝟒 𝐣 , ⃑𝐑 ⃑ = −𝟑𝐢 − 𝟒 𝐣 𝐚𝐜𝐭𝐬 𝐚𝐭 𝐂 𝐟𝐢𝐧𝐝 ⃑⃑⃑ 𝐅𝟐 𝐚𝐧𝐝 𝐁𝐂 ⃑⃑ = 𝑭 ⃑⃑⃑⃑𝟏 + 𝑭 ⃑⃑⃑⃑𝟐 so 𝑭 ⃑⃑⃑⃑𝟐 = 𝑹 ⃑⃑ − 𝑭 ⃑⃑⃑⃑𝟏 = −𝟔𝒊 − 𝟖 𝒋 𝑹 ⃑⃑⃑⃑𝟏 ‖ = √𝟑𝟐 + 𝟒𝟐 = 𝟓 ‖𝑭 𝑭𝟏 𝑭𝟐 = 𝑪𝑩 𝑪𝑨 so 𝟓 𝟏𝟎 = 𝑪𝑩 𝑪𝑩+𝟐𝟎 𝑭𝟐 ‖ = √(−𝟔)𝟐 + (−𝟖)𝟐 = 𝟏𝟎 ‖ ⃑⃑⃑⃑ 𝒕𝒉𝒆𝒏 𝑪𝑩 = 𝟐𝟎𝒄𝒎 (𝟖)𝐓𝐰𝐨 𝐩𝐚𝐫𝐚𝐥𝐥𝐞𝐥 𝐟𝐨𝐫𝐜𝐞𝐬 ⃑⃑⃑ 𝐅𝟏 𝐚𝐧𝐝 ⃑⃑⃑ 𝐅𝟐 𝐚𝐜𝐭 𝐚𝐭 𝐀(𝟏, 𝟑)𝐚𝐧𝐝𝐁(𝟒, 𝟗) 𝐫𝐞𝐬𝐩𝐞𝐜𝐭𝐢𝐯𝐞𝐥𝐲 ⃑⃑⃑ 𝐅𝟏 = 𝟐𝐢 − 𝟑 𝐣 , ⃑⃑⃑𝟐 = 𝟒𝐢 − 𝟔 𝐣 𝐅 𝐟𝐢𝐧𝐝 𝐭𝐡𝐞 𝐫𝐞𝐬𝐮𝐥𝐭𝐚𝐧𝐭 , 𝐭𝐡𝐞 𝐞𝐪𝐮𝐚𝐭𝐢𝐨𝐧 𝐨𝐟 𝐭𝐡𝐞 𝐥𝐢𝐧𝐞 𝐨𝐟 𝐚𝐜𝐭𝐢𝐨𝐧 𝐨𝐟 𝐭𝐡𝐞 𝐫𝐞𝐬𝐮𝐥𝐭𝐚𝐧𝐭 𝐚𝐧𝐝 𝐢𝐭𝐬 𝐩𝐨𝐢𝐧𝐭 𝐨𝐟 𝐚𝐜𝐭𝐢𝐨𝐧 ⃑⃑ = 𝑭 ⃑⃑⃑⃑𝟏 + 𝑭 ⃑⃑⃑⃑𝟐 = 𝟔𝒊 − 𝟗 𝒋 acts at C(x,y) 𝑹 24 ⃑ ‖ = √𝟔𝟐 + (−𝟗)𝟐 = ξ𝟏𝟏𝟕 ‖ ⃑𝑹 we may get the point of action of the resultant by getting the intersection of the line of action of resultant and the line passes by the two points A and B at first get the equation of the line of action of resultant 𝒍𝒆𝒕 𝒕𝒉𝒆 𝒓𝒆𝒔𝒖𝒍𝒕𝒂𝒏𝒕𝒕 𝒂𝒄𝒕𝒔 𝒂𝒕 𝑪(𝒙, 𝒚) ⃑⃑⃑⃑⃑⃑ 𝑿𝑭 ⃑⃑⃑⃑⃑⃑ 𝑿𝑭 ⃑⃑⃑⃑⃑⃑ 𝑿𝑹 ⃑⃑⃑⃑⃑⃑ ⃑⃑⃑⃑𝟏 + 𝑶𝑩 ⃑⃑⃑⃑𝟐 = 𝑶𝑪 ⃑⃑ 𝑴𝑶 = 𝑶𝑨 (𝟏, 𝟑)𝑿(𝟐, −𝟑) + (𝟒, 𝟗)𝑿(𝟒, −𝟔) = (𝒙, 𝒚)𝑿(𝟔, −𝟗) −𝟔𝟗 = −𝟗𝒙 − 𝟔𝒚 𝒕𝒉𝒆𝒏 𝟑𝒙 + 𝟐𝒚 = 𝟐𝟑 second get the equation of the line joining A and B A(1,3) , B(4,9) 𝒎= 𝟗−𝟑 =𝟐 𝟒−𝟏 equation of the line y-3=2[x-1] so y=2x+1 Third Now solve these two equations together 3x+2y=23 the equation of the resultant and y=2x+1 the equation of the line joining A and B 3x+2[2x+1]=23 3x+4x+2=23 so x=3 then y=7 the point (3,7) 9] R=50N, AB=45 cm, BC=30 cm find F1,F2 𝑅 = 𝐹1 − 𝐹2 = 50 → (1) 𝐹1 𝐴𝐶 𝐹1 75 5 = → = = 𝐹2 𝐴𝐵 𝐹2 ∴ 3𝐹1 = 5𝐹2 5 → 45 3 5 ∴ 𝐹1 = 𝐹2 3 2 ∴ 𝑖𝑛 (1) 𝐹2 − 𝐹2 = 50 → 𝐹 = 50 3 3 2 ∴ 𝐹1 − 75 = 50 → 𝐹1 = 125𝑁 → ∴ 𝐹2 = 75𝑁 25 10)A motorcycle of mass 200kg and its weight acts at the vertical line passing through the midpoint between the two wheels, and the mass of the motorcyclist is 84 kg. and its weight acts at the vertical line which is distant 1 meter behind the front wheel.Find the reaction of the ground on each of the two wheels 𝑹𝟏 + 𝑹𝟐 = 𝟐𝟖𝟒 𝑴𝑨 = [−𝟐𝟎𝟎][𝟕𝟎] + [−𝟖𝟒][𝟏𝟎𝟎] + [𝑹𝟐 ][𝟏𝟒𝟎] = 𝟎 𝑹𝟐 = 𝟏𝟔𝟎 𝒌𝒈. 𝒘𝒕 , 𝑹𝟏 = 𝟏𝟐𝟒𝒌𝒈. 𝒘𝒕 R1 R2 C A D 30 70 200 B 40 84 11)An iron beam of length 50 cm and weight 75Newton acting at its midpoint rests on two supports the distance between the two supports is 24 cm. if the pressure on one of the supports is twice the pressure on the other then Find the distance between each support and the end nearer to it 50 r 2r L 24 A Let AC = L R = 0  r + 2r – 75 = 0  3r = 75  r = 25 MA = 0 B C ( rL ) + ( -75  25 ) + ( 2r ) ( L + 24 ) = 0 25L – 1875 + 50 L + 1200 = 0 75 L = 675  L = 9 AC = 9cm , BD = 50 – ( 24 + 9 ) = 50 – 33 = 17cm. D 7 5 26 12) AB is a rod of length 50cm and weight 10 Newton acting at its midpoint rests in a horizontal position on two supports one of them at a point 15cm from A , and the other 10 cm from B. What is the mag. Of the weight, should be suspended from B such that the rod is on the point of rotation. [about to rotate about E][ about to separate from C] r1 + r2 –10 = 0 → MC = 0 r1 + r2 = 10 r1  -10  10 + r2  25 = 0 r2 = 4 r1 = 6 r2 25 A D 15 10 25 15 10 C B E 2. 10 W the rod is about to rotate about E  r1 = 0 r2 = w + 10 ME = 0 ( -W W = 15 Newton  10 ) + ( 10  15 ) = 0 r2 = 25 Newton 13) A uniform rod of length 80 cm and weight 3N is suspended by two vertical strings from its ends if each string can bear a maximum tension of 5 N Find a point such that a weight 4N suspended such that the nearer string is about to be cut 5 T L 40-L A 40 B 4 3 The nearer string is about to be cut so the tension in it is 5N T=2 MA=(-4L)+(-3X40)+(2X80)=0 L=10 its 10 cm from the nearer end 27 14)ABCD is a non uniform rod rests horizontally on two smooth supports at B and C such that AB=6cm and CD=7 cm, and the point of action of the rod weight divides it internally by the ratio 2:3 from A, its found that if a weight 120gm.wt is suspended from A or a weight 180 gm.wt is suspended from D,the rod is about to rotate.Find the weight of the rod and the distance between the two supports. R1 R2 A 6 2L-6 3L-7 D 7 D c C B 2L 120 7 W 3L 𝑴𝑩 = [𝟏𝟐𝟎][𝟔] + [−𝑾][𝟐𝑳 − 𝟔] = 𝟎 𝒔𝒐 [𝒘][𝟐𝑳 − 𝟔] = 𝟕𝟐𝟎 R1 R2 A 2L-6 6 3L-7 c C B 2L W 3L 𝑴𝑪 = [𝒘][𝟑𝑳 − 𝟕] + [−𝟏𝟖𝟎][𝟕] = 𝟎 𝒔𝒐 [𝒘][𝟑𝑳 − 𝟕] = 𝟏𝟐𝟔𝟎 [𝒘][𝟑𝑳 − 𝟕] 𝟏𝟐𝟔𝟎 𝟑𝑳 − 𝟕 𝟕 = 𝒔𝒐 = [𝒘][𝟐𝑳 − 𝟔] 𝟕𝟐𝟎 𝟐𝑳 − 𝟔 𝟒 𝟏𝟐𝑳 − 𝟐𝟖 = 𝟏𝟒𝑳 − 𝟒𝟐 𝟐𝑳 = 𝟏𝟒 𝒕𝒉𝒆𝒏 𝑳 = 𝟕 𝒂𝒏𝒅 𝑾 = 𝟗𝟎 𝒈𝒎. 𝒘𝒕 the distance between the two supports is 5L-13=22 cm 180 28 15) AB is heavy un uniform rod of length 140cm is suspended by two vertical strings one of them at B the other is 40 cm from A. If the tension in the string at B equals 1 the tension in the other 4 string. Find the point of action of the weight. if the greatest weight suspended for A equals 12 N. Find the weight of the rod. 140 T2 T1 40 x A B L 1 T1 = T2  T2 = 4T 4 W T1 = T T1 + T2 – w=0  W=5T MA = 0 40 T2 – w ( Ax ) + 140 T1 = 0 160T – 5T ( Ax ) + 140 T = 0 160 – 5Ax + 140 = 0 5( Ax ) = 300 T2 Ax = 60cm T1 = 0 T2 = w + 12 MC = 12  40 – w ( 20 ) = 0 A 40 D B C 60 12 W w = 24 16)AB is a non-uniform wooden board of length 4 meters rests horizontal on two supports at C and D such that AC=1m and BD=1.5m,if the maximum distance a 780 Newton man can move on the board from A to B without getting the board unbalanced is 3m,and the maximum distance the same man can move from B to A is 3.5m find the weight of the board and its point of action. 29 the rod is about to overturn about D so the reaction at C=0 𝑴𝑫 = [𝟕𝟖𝟎][𝟎. 𝟓] + [−𝒘][𝟐. 𝟓 − 𝒙] = 𝟎 𝒔𝒐 [𝒘][𝟐. 𝟓 − 𝒙] = 𝟑𝟗𝟎 the rod is about to overturn about C then the reaction at D=0 𝑴𝑪 = [−𝟕𝟖𝟎][𝟎. 𝟓] + [𝒘][𝒙 − 𝟏] = 𝟎 𝒔𝒐 [𝒘][𝒙 − 𝟏] = 𝟑𝟗𝟎 [𝒘][𝟐. 𝟓 − 𝒙] = [𝒘][𝒙 − 𝟏] [𝟐. 𝟓 − 𝒙] = [𝒙 − 𝟏] 𝒔𝒐 𝟐𝒙 = 𝟑. 𝟓 𝒕𝒉𝒆𝒏 𝒙 = 𝟏. 𝟕𝟓 𝒂𝒏𝒅 𝒘 = 𝟓𝟐𝟎𝑵𝒆𝒘𝒕𝒐𝒏 30 Friction 1) A body of mass 80 gms rests on a rough horizontal plane If the body is about to move under the action of a horizontal force 20 gm. Wt. then the coefficient of friction=………………………. the least force inclines to the plane upward by 𝜶 so that 𝒄𝒐𝒔 𝜶 = the magnitude of the resultant reaction =…………… that direction of the resultant reaction =…………… when the mass is 80 gram then the weight is 80 gm.wt The body is about to move \Friction is limiting and equals 𝝁𝒔 𝑵 ∑ 𝑿 = 𝟎 → 𝑭 − 𝝁𝒔 𝑵 = 𝟎 → 𝝁𝒔 𝑵 = 𝑭 = 𝟐𝟎 𝒈𝒎. 𝒘𝒕 ∑ 𝒀 = 𝟎 → 𝑵 − 𝟖𝟎 = 𝟎 → 𝑵 = 𝟖𝟎 𝒈𝒎. 𝒘𝒕 𝒔𝒐 𝝁𝒔 = 𝒇 𝑵 = 𝟐𝟎 𝟖𝟎 ∑ 𝑿 = 𝟎 → 𝑭𝒄𝒐𝒔𝜶 − 𝝁𝒔 𝑵 = 𝟎 𝟒𝑭 𝟏 − 𝑵 = 𝟎 … … … . (𝟏) 𝟓 𝟒 ∑ 𝒀 = 𝟎 → 𝒇 𝒔𝒊𝒏𝜶 + 𝑵 − 𝟖𝟎 = 𝟎 𝟑𝑭 𝟓 + 𝑵 = 𝟖𝟎 𝟑 𝑵 = 𝟖𝟎 − 𝒇 … . (𝟐) 𝟓 𝒇𝒓𝒐𝒎(𝟏)𝒂𝒏𝒅 (𝟐) 𝟒 𝟑 𝟏𝟗 𝟒𝟎𝟎 𝒇 − 𝟐𝟎 + 𝒇 = 𝟎 𝒔𝒐 𝒇 = 𝟐𝟎 → 𝒇 = 𝒈𝒎. 𝒘𝒕 , 𝟓 𝟐𝟎 𝟐𝟎 𝟏𝟗 𝟒𝟎𝟎 𝟑 𝟏𝟐𝟖𝟎 𝑵 = 𝟖𝟎 − × = 𝒈𝒎. 𝒘𝒕 𝟏𝟗 𝟓 𝟏𝟗 𝑹𝒕 = 𝑵√[𝝁𝟐 + 𝟏] = 𝟏𝟐𝟖𝟎 𝟏 √ + 𝟏 = 𝟔𝟗. 𝟒𝟒𝟐 𝒈𝒎. 𝒘𝒕 𝟏𝟗 𝟏𝟔 = 𝟏 𝟒 𝟒 𝟓 that would move the body=…. 31 𝒕𝒂𝒏 𝝀 = 𝝁𝒔 = 𝟏 𝟏 𝒔𝒐 𝝀 = 𝒕𝒂𝒏−𝟏 ( ) = 𝟏𝟒. 𝟎𝟒𝟎 𝟒 𝟒 2) A body of mass 8gm is placed on a rough horizontal plane, two horizontal forces 2,𝟐 gm.wt include an angle of measure 1200 act upon the body if the body is at rest Then that the coefficient 𝝁𝒔 ≥……………………. 𝑭 = √𝑭𝟐 𝟏 + 𝑭𝟐 𝟐 + 𝟐𝑭𝟏 𝑭𝟐 𝒄𝒐𝒔𝜶 = √𝟒 + 𝟒 + 𝟐[𝟐][𝟐][ −𝟏 ] = 𝟐 𝒈𝒎. 𝒘𝒕 𝟐 𝑭 = 𝑭𝒓 = 𝟐 𝒂𝒏𝒅 𝒓 = 𝟖 𝑭𝒓 ≤ 𝝁𝒔 𝒓 𝒔𝒐 𝝁𝒔 ≥ 𝟐 ≤ 𝟖𝝁𝒔 𝟏 𝟒 3) A body of weight W is placed on a rough horizontal plane two perpendicular horizontal forces 𝑭𝟏 𝒂𝒏𝒅 𝑭𝟐 act upon the body if the body is in equilibrium prove that the coefficient of friction 𝝁𝒔 ≥ √𝑭𝟏 𝟐 +𝑭𝟐 𝟐 𝑾 𝑭 = √𝑭 𝟐 𝟏 + 𝑭 𝟐 𝟐 𝑭 ≤ 𝝁𝒔 𝒓 𝑭 ≤ 𝝁𝒔 𝑾 √𝑭𝟐 𝟏 + 𝑭𝟐 𝟐 ≤ 𝝁𝒔 𝑾 √𝑭𝟐 𝟏 + 𝑭𝟐 𝟐 𝝁𝒔 ≥ 𝑾 32 4)A body of weight 8N on a horizontal plane and the coef. Of friction between the body and the plane is 0.5 then the interval by which the force of friction belongs is ……………. the interval by which the resultant reaction belongs is ……………. 𝑭𝒓 𝝐[𝟎, 𝝁𝒔 𝑵] 𝒔𝒐 𝑭𝒓 𝝐[𝟎, 𝟒] but 𝑹𝒕 = √𝒓𝟐 + 𝑭𝒓 𝟐 𝒕𝒉𝒆𝒏 𝑹𝒕 𝝐[√𝒓𝟐 + 𝟎𝟐 , √𝒓𝟐 + 𝟒𝟐 ] 𝑹𝒕 𝝐[𝟖, ξ𝟖𝟎] 5) A body of weight W is placed on a rough plane the measure of the angle of friction is 𝝀 Find the magnitude and the direction of the least force would move the body given that the plane is horizontal Let the force inclines to plane by angle 𝜽 Since the body is about to move so 𝑭𝒓 = 𝑭𝑺 = 𝝁𝒔 𝒓 Let Rt to be the resultant of r and 𝝁𝒔 𝒓 Using Lami’s rule 𝑭 𝑾 𝑾𝒔𝒊𝒏𝝀 = 𝒔𝒐 𝑭 = 𝐬𝐢𝐧 [𝟏𝟖𝟎 − 𝝀] 𝐬𝐢𝐧 [𝟗𝟎 − 𝜽 + 𝝀] 𝐜𝐨𝐬 [𝜽 − 𝝀] 𝐑𝐞𝐦𝐞𝐦𝐛𝐞𝐫 𝐭𝐡𝐚𝐭 𝐬𝐢𝐧[𝟗𝟎 − 𝜽] = 𝒄𝒐𝒔 𝜽 The least force when 𝐜𝐨𝐬[𝜽 − 𝝀] 𝒊𝒔 𝒎𝒂𝒙𝒊𝒎𝒖𝒎 which equals 1 That happens when 𝜽 − 𝝀 = 𝟎 𝒎𝒆𝒂𝒏𝒔 𝒘𝒉𝒆𝒏 𝜽 = 𝝀 𝒂𝒏𝒅 𝒕𝒉𝒆 𝒇𝒐𝒓𝒄𝒆 𝒕𝒉𝒆𝒏 𝑭 = 𝒘 𝒔𝒊𝒏 𝝀 6) A body whose weight is 4 Kg .wt is placed on a rough inclined plane inclines by 600 to the horizontal and the coefficient of friction between the body and the plane is 3 2 Determine whether the body is sliding or about to slide or the friction is not limiting 𝜽 = 𝟔𝟎 , 𝒃𝒖𝒕 𝝁𝒔 = 𝒕𝒂𝒏 𝝀 = ξ𝟑 𝟐 ⇒ 𝝀 = 𝟒𝟎. 𝟖𝟗 33 𝜽 > 𝝀 𝒔𝒍𝒊𝒅𝒊𝒏𝒈 7) A body Of weight 36 Newton is placed on the point of sliding down on an inclined plane inclines by 600 to the horizontal under the action of its weight only If the Inclination is decreased to 30° find the friction force Then find the force that acts upon the body in the direction parallel to the line of the greatest slope to make the body about to move i) up the plane ii) Down the plane ∵The body is about to slide down under the action of its weight only 𝛂 = 𝟔𝟎0 ∴ 𝛌 = 𝛂 = 𝟔𝟎𝟎 and 𝝁𝒔 = 𝐭𝐚𝐧𝟔𝟎 = ξ𝟑 when the inclination is decreased to 300 so the body comes to rest and friction becomes not limiting and equals Fr From equilibrium ∑ 𝐲 = 𝟎 ⇒ 𝐍 = 𝟑𝟔𝐜𝐨𝐬𝟑𝟎 = 𝟏𝟖ξ𝟑𝑵 ∑ 𝐱 = 𝟎 ⇒ 𝐅𝐫 = 𝟑𝟔𝐬𝐢𝐧𝟑𝟎 = 𝟏𝟖𝐍 The least force that would move the body up the plane will be up the plane Since the body is about to move Friction is limiting 𝝁𝒓 𝒂𝒏𝒅 𝒂𝒄𝒕𝒔 𝒅𝒐𝒘𝒏 𝒕𝒉𝒆 𝒑𝒍𝒂𝒏𝒆 𝑵 = 𝟑𝟔 𝒄𝒐𝒔 𝟑𝟎 = 𝟏𝟖ξ𝟑 𝑭 = 𝟑𝟔 𝒔𝒊𝒏 𝟑𝟎 + 𝝁𝒔 𝑵 = 𝟕𝟐𝑵 34 The least force that would move the body down the plane will be down the plane Since the body is about to move Friction is limiting = 𝝁𝒔 𝐍 𝐚𝐧𝐝 𝐚𝐜𝐭𝐬 𝐮𝐩 𝐭𝐡𝐞 𝐩𝐥𝐚𝐧𝐞 𝑵 = 𝟑𝟔 𝒄𝒐𝒔 𝟑𝟎 = 𝟏𝟖ξ𝟑 𝑭 + 𝟑𝟔 𝒔𝒊𝒏 𝟑𝟎 = 𝝁𝒔 𝑵 𝑭 = 𝟑𝟔Newton 8) A body of mass 5 kg is placed on a rough inclined plane under the action of a force F parallel to the plane upwards it was found that if F=2kg .wt the body is about to move down ward and if F=3 kg.wt, then the body is about to move up the plane so find the coef. Of friction between the body and the plane as well as the measure of the angle of inclination of the plane to the horizontal. First case when the force is 2 then its about to move down 𝑵 = 𝟓 𝒄𝒐𝒔 𝜶 𝟓 𝒔𝒊𝒏𝜶 = 𝟐 + 𝝁𝒔 𝑵 𝒔𝒐 𝟓 𝒔𝒊𝒏𝜶 − 𝝁𝒔 [𝟓 𝒄𝒐𝒔 ∝] = 𝟐______(𝟏) The second case when the force is 3 then the body is about to move upwards 𝑵 = 𝟓 𝒄𝒐𝒔 𝜶 𝟓 𝒔𝒊𝒏𝜶 + 𝝁𝒔 𝑵 = 𝟑 𝟓 𝒔𝒊𝒏𝜶 + 𝝁𝒔 [ 𝟓 𝒄𝒐𝒔 𝜶] = 𝟑______(𝟐) from (1) and (2) by adding 𝟏𝟎 𝒔𝒊𝒏𝜶 = 𝟓 𝒔𝒐 𝒔𝒊𝒏 ∝= 𝟏 𝒕𝒉𝒆𝒏 ∝= 𝟑𝟎𝟎 𝟐 𝒃𝒚 𝒓𝒆𝒑𝒍𝒂𝒄𝒆𝒎𝒆𝒏𝒕 𝒊𝒏 𝒆𝒒𝒖𝒂𝒕𝒊𝒐𝒏 (𝟏) 𝟓 𝒔𝒊𝒏𝟑𝟎 − 𝟓𝝁𝒔 𝒄𝒐𝒔𝟑𝟎 = 𝟐 35 𝟓 𝟓ξ 𝟑 − 𝝁 =𝟐 𝟐 𝟐 𝒔 , 𝟓 − 𝟓ξ𝟑𝝁𝒔 = 𝟒 , 𝝁𝒔 = 𝟏 𝟓ξ 𝟑 = ξ𝟑 𝟏𝟓 9) A body of mass 130 kg is placed on a rough inclined plane which inclines to the horizontal by an angle whose cosine 𝟓 𝟏𝟑 coefficient Of friction is under the action of a force F parallel to the plane upwards if the 𝟐 𝟓 Find the limits between the force lies to make the body at rest The body is about to slide down 𝒓 = 𝟏𝟑𝟎 𝒄𝒐𝒔𝜶 = 𝟓𝟎 𝑭 = 𝟏𝟑𝟎𝒔𝒊𝒏𝜶 − 𝝁𝒓 = 𝟏𝟎𝟎 The body is about to move up 𝒓 = 𝟏𝟑𝟎 𝒄𝒐𝒔𝜶 = 𝟓𝟎 𝑭 = 𝟏𝟑𝟎𝒔𝒊𝒏𝜶 + 𝝁𝒓 = 𝟏𝟒𝟎 𝟏𝟎𝟎 ≤ 𝑭 ≤ 𝟏𝟒𝟎 10)Two bodies of weights 3w,4w connected together with a light string coincide with the line of the greatest slope of an inclined rough plane the coefficient of the static friction between each of the first and the second with the plane is 𝟏 𝟔 𝒂𝒏𝒅 𝟏 𝟖 respectively. If we begin to tilt the plane gradually which body should be put below the other such that the two bodies will move together keeping the string tout then find the angle of inclination at this instant. 𝟒𝒘 𝒔𝒊𝒏 ∝= 𝟏 𝒓 + 𝑻 , 𝒓𝟏 = 𝟒𝒘 𝒄𝒐𝒔 ∝ 𝟖 𝟏 36 𝒔𝒐 𝟒𝒘 𝒔𝒊𝒏 ∝= 𝟏 (𝟒𝒘 𝒄𝒐𝒔 ∝ ) + 𝑻 𝒕𝒉𝒆𝒏 𝟖 𝟏 𝑻 = 𝟒𝒘 𝒔𝒊𝒏 ∝ − 𝒘 𝒄𝒐𝒔 ∝ ____(𝟏) 𝟐 𝟏 𝒓 , 𝒓𝟐 = 𝟑𝒘 𝒄𝒐𝒔 ∝ , 𝟔 𝟐 𝟏 𝟏 𝑻 + 𝟑𝒘 𝒔𝒊𝒏 ∝= (𝟑𝒘 𝒄𝒐𝒔 ∝ ) 𝒕𝒉𝒆𝒏 𝑻 = 𝒘 𝒄𝒐𝒔 ∝ −𝟑𝒘 𝒔𝒊𝒏 ∝ ______(𝟐) 𝟔 𝟐 𝑻 + 𝟑𝒘 𝒔𝒊𝒏 ∝= from (1) and (2) 𝟏 𝟏 𝟒𝒘 𝒔𝒊𝒏 ∝ − 𝒘 𝒄𝒐𝒔 ∝= 𝒘 𝒄𝒐𝒔 ∝ −𝟑𝒘 𝒔𝒊𝒏 ∝ 𝟐 𝟐 𝟕𝒘 𝒔𝒊𝒏 ∝= 𝒘 𝒄𝒐𝒔 ∝ 𝒔𝒊𝒏 ∝ 𝟏 = 𝒄𝒐𝒔 ∝ 𝟕 𝒔𝒐 𝑻𝒂𝒏 𝜶 = 𝟏 𝟕 𝒔𝒐 𝜶 = 𝒕𝒂𝒏−𝟏 𝟏 = 𝟖. 𝟏𝟑𝟎 𝟕 11) A body of weight W is placed on a rough plane the measure of the angle of friction is 𝝀 find the magnitude and the direction of the least force that would move the body up given that the plane is inclined to the horizontal by ∝. Let the force acts up wards and inclines by θ to the plane 𝒘 𝑭 = 𝐬𝐢𝐧(𝟗𝟎 − 𝝑 + 𝝀) 𝐬𝐢𝐧(𝟏𝟖𝟎 − 𝜶 − 𝝀) 𝒘 𝑭 = 𝐜𝐨𝐬(𝝑 − 𝝀) 𝐬𝐢𝐧(𝜶 + 𝝀) 𝑭= 𝒘𝒔𝒊𝒏(𝜶 + 𝝀) 𝐜𝐨𝐬(𝝑 − 𝝀) 𝒕𝒉𝒆 𝒍𝒆𝒂𝒔𝒕 𝒇𝒐𝒓𝒄𝒆 𝒐𝒇 𝑭 𝒊𝒔 𝒂𝒕 𝒎𝒂𝒙. 𝒐𝒇 𝐜𝐨𝐬(𝝑 − 𝝀) 𝒘𝒉𝒊𝒄𝒉 𝒆𝒒𝒖𝒂𝒍𝒔 𝟏 𝒂𝒕 𝝑 − 𝝀 = 𝟎 𝒐𝒓 𝝑 = 𝝀 𝒔𝒐 𝒕𝒉𝒆 𝒍𝒆𝒂𝒔𝒕 𝒐𝒇 𝑭 = 𝒘 𝐬𝐢𝐧 (𝜶 + 𝝀) 37 The general equilibrium 1)A uniform bar AB is hinged at A to a vertical wall, its weight 10N and 200 cm long, A weight of 10Newton is hung from B and the bar is kept horizontally by a string attached to a point 150 cm from A and the string is fixed to the wall at a point above A if the string is inclined to the horizontal by 300 . find the tension in the string and the reaction of the hinge. =0 =0 2)A uniform rod of weight W is hinged at an end ,the other end is attached to a string joined to a point at the same horizontal plane passing through the hinges such that the measure of the angle of inclination for each of the rod and the string to the horizontal is 𝜽.given that the length of the rod is L Prove that the reaction at the hinge is equal to 𝒘√𝒄𝒐𝒕𝟐 𝜽+𝟗 x+T cos 𝜽=0 so x=-T cos 𝜽 Y+T sin 𝜽-W=0 then Y=w-T sin 𝜽 𝟏 𝑴𝑨 = [−𝒘] [ 𝑳 𝒄𝒐𝒔𝜽] + [𝑻 𝒔𝒊𝒏𝜽][𝟐𝑳 𝒄𝒐𝒔𝜽] = 𝟎 𝟐 divide by 𝟏 𝟐 𝑳 𝒄𝒐𝒔𝜽 𝒘 𝟒𝒔𝒊𝒏𝜽 𝒘 𝒘 𝟑𝒘 𝒀=𝒘− 𝒔𝒊𝒏𝜽 = 𝒘 − = 𝟒𝒔𝒊𝒏𝜽 𝟒 𝟒 −𝒘 + 𝟒𝑻 𝒔𝒊𝒏𝜽 = 𝟎 𝒔𝒐 𝒘 = 𝟒𝑻 𝒔𝒊𝒏𝜽 𝒔𝒐 𝑻 = 𝒙= −𝒘𝒄𝒐𝒔𝜽 , 𝟒𝒔𝒊𝒏𝜽 𝟒 38 𝑹= √𝒙𝟐 + 𝒚𝟐 −𝒘𝒄𝒐𝒔𝜽 𝟐 𝟑𝒘 𝟐 𝒘𝟐 𝒄𝒐𝒔𝟐 𝜽 𝟗𝒘𝟐 √ √ = [ + ] +[ ] = 𝟒𝒔𝒊𝒏𝜽 𝟒 𝟏𝟔 𝒔𝒊𝒏𝟐 𝜽 𝟏𝟔 𝒘𝟐 𝒄𝒐𝒔𝟐 𝜽 𝒘 √𝒄𝒐𝒕𝟐 𝜽 + 𝟗 =√ [ + 𝟗] = 𝟏𝟔 𝒔𝒊𝒏𝟐 𝜽 𝟒 3)A uniform rod of weight 40N rests in a vertical plane with one end against a rough vertical wall with one end against rough horizontal floor if the coef. Of frictions between the rod and each of the wall and the floor are 𝟏 𝟐 𝒂𝒏𝒅 𝟏 𝟑 respectively, the rod inclines to the horizontal by 450 Find the least horizontal force that will make the lower end of the rod about to move towards the wall. Let the length is 2L 1 1 r1 + r2 - f = 0 Þ f = r1 + r2 3 3 1 r2 - r1 = 40 2 M B = 40[L sin 45] 1 + r1[2L sin 45] 2 - r1[2L cos 45] = 0 40 + r1 - 2r1 = 0 Þ r1 = 40 , r2 = 60 , f = 60N 4) A uniform rod rests in a vertical plane with the upper end against a smooth vertical wall and the lower end on a horizontal rough ground the coefficient of friction between the rod and the 𝟏 ground is if the rod rests in limiting equilibrium find the angel of inclination of the rod to the 𝟑 wall. 39 Let the length is 2L 1 Rx = r1 - mr2 = 0 Þ r1 = r2 3 1 Ry = r2 - w = 0 Þ r2 = w Þ r1 = W 3 M B = w[L sinq ] - r1[2L cosq ] = 0 1 w[L sinq ] = [ W ][2L cosq ] 3 sinq 2 2 = Þ Tanq = cosq 3 3 θ 𝜇2𝑅2 𝜽 = 𝟑𝟑. 𝟕𝟎 5)A uniform ladder of weight 20kg.wt rests in a vertical plane with one end on a rough horizontal ground and the other end against a smooth vertical wall , the ladder is inclined by 60 0 to the horizontal given that the coef. Of friction between the ladder and the ground equals 𝟏 𝟐ξ𝟑 Prove that the maximum distance a man of mass 60 kg can ascend equals half the length of the ladder Let the length is 2L Rx = r2 - mr1 = 0 Þ r2 = 1 2 3 r1 40 3 M A = 20[L sin 30]+ 60[xsin 30]- r2 [2L sin60] = 0 Ry = r1 - 80 = 0 Þ r1 = 80 Þ r2 = æ 40 ö 10L + 30x - 3L ç ÷ = 0 è 3ø 10L + 30x - 40L = 0 30x = 30L Þ x = L 40 6)A uniform ladder of weight 30Kg wt rests with one end against a smooth vertical wall and with the other end on a rough horizontal ground such that it is inclined to it by 450. If a man 60Kg wt ascends that ladder slowly so that the ladder becomes on the point of sliding when the man has gone up to ¾ the length of the ladder so prove that the coefficient of friction between the ground and the ladder is 2/3. If the man wants to reach the top of the ladder so find the least horizontal force which acts at the lowest end of the ladder so that the man may do so. Also the weight may replace the force at the lower end to support the body Given that the coef. Of friction between the body and the ground is 0.9 Let the length of the ladder be 4L the end A is about to slide away from the wall. So friction at is limiting =µ r ∑ 𝑿 = 𝟎 → 𝒓₂ = µ 𝒓₁ (1) ∑ 𝒀 = 𝟎 → 𝒓₁ = 𝟗𝟎 (2) Moment with respect to A = 0 -30 × 2L cos45⁰ - 60 × 3L cos45⁰ + r₂ × 4L sin 45⁰ = 0 Divided by 2L cos 450 -15 - 45 + r= 0 r = 60 In (1) 60 = µ 90 µ = 2/3 Let the least force at A be F by which the man reaches B when the ladder is about to slide away from the wall. So friction at A is limiting and equals µ r From equilibrium ∑ 𝑿 = 𝟎 → 𝒓𝟐 = ∑𝑿 = 𝟎 → 𝟐 𝟑 𝒓₁ + 𝑭 𝒓₁ = 𝟗𝟎 (1) (2) Moment with respect to A = 0 -30 × 2L cos45⁰ -60 × 4L cos45⁰ + r₂ × 4L sin 45⁰ = 0 Divided by 4L cos45⁰ -15 – 60 + r₂ = 0 → r₂ = 75 41 In (1) 75 = 2/3 × 90 + F → F = 15kg.wt 𝒕𝒉𝒆 𝒘𝒆𝒊𝒈𝒉𝒕 𝒕𝒉𝒂𝒕 𝒓𝒆𝒑𝒍𝒂𝒄𝒆 𝒕𝒉𝒆 𝒇𝒐𝒓𝒄𝒆 𝑭 𝑭 = 𝝁𝟑 𝑾 𝒔𝒐 𝑾 = 𝑭 𝟏𝟓 = = 𝟏𝟔. 𝟔𝒌𝒈. 𝒘𝒕 𝝁 (𝟎. 𝟗) 7) A uniform ladder of weight “w” rests with one end on a rough vertical wall and with its other end on a rough horizontal floor. If “µ1”,”µ2” are the coefficients of friction between the ladder and the wall and between the ladder and the floor respectively, and the ladder was in a state of limiting equilibrium making an angle of measure Ɵ with the vertical wall. Prove that 𝑻𝒂𝒏 𝜽 = 𝟐µ₂ 𝟏−𝝁₁µ₂ 𝒊𝒇 𝝁𝟏 = 𝝁𝟐 𝒕𝒉𝒆𝒏 𝜽 = 𝟐𝝀 𝒈𝒊𝒗𝒆𝒏 𝒕𝒉𝒂𝒕 𝝀 𝒊𝒔 𝒕𝒉𝒆 𝒂𝒏𝒈𝒍𝒆 𝒐𝒇 𝒇𝒓𝒊𝒄𝒕𝒊𝒐𝒏 r1 - m2r2 = 0 r2 + m1r1 - w = 0 M C = ( -W ) ( Lsinq ) + r2 ( 2Lsinq ) - r1 ( 2Lcosq ) = 0 ¸Lcosq -wTanq + 2r2Tanq = 2r Tanq = 2r1 2 m2r2 2 m2r2 2m = = = 2 2r2 -W 2r2 - r2 - m1r1 2r2 - r2 - m1 ( m2r2 ) 1- m1m2 𝒕𝒂𝒏 𝜽 = 𝟐 𝒕𝒂𝒏𝝀 = 𝒕𝒂𝒏 𝟐𝝀 𝒔𝒐 𝜽 = 𝟐𝝀 𝟏 − 𝒕𝒂𝒏𝟐 𝝀 42 8) A smooth hollow hemisphere of radius r rests with its circular base on a rough horizontal ground. A uniform rod of length 2r and weights W is placed with one of its points touching the spherical surface at B and resting with one on the ground. If the rod becomes in limiting equilibrium when AB = 3/2 r , find the reaction Between the spherical surface and the rod and then prove that the coefficient of friction between the ground and the rod is 4/7 Solution R\ cos 𝛼 In ∆ ABO : (∠𝑶𝑨𝑩) = 𝜶 R\ sin 𝛼 AB = 3/2 r , OB = r tan α= 𝑶𝑩 𝑨𝑩 = 𝒓 𝟑 𝒓 𝟐 = ∝ 𝟐 𝟑 ∵ The rod is about to slide so friction is limiting = µ r by resolving the forces in the directions of X and Y axes. ∑ 𝑿 = 𝟎 → 𝑹′ 𝐬𝐢𝐧 𝜶 = µ 𝑹 → 𝑹′ 𝟐 ξ𝟏𝟑 ∑ 𝒀 = 𝟎 → 𝑹′ 𝐜𝐨𝐬 𝜶 + 𝑹 = 𝑾 → 𝑹′ =µ𝑹 𝟑 ξ𝟏𝟑 +𝑹=𝑾 (1) (2) Moments with respect to A = 0 𝟑 𝟑 𝟐 ξ𝟏𝟑 -R’ 𝒓 + 𝑾 𝒓 𝐜𝐨𝐬 𝜶 = 𝟎 → 𝑾 𝒓 So 𝑹′ = In (2) In (1) 𝟐 ξ𝟏𝟑 𝟐 ξ𝟏𝟑 𝟐 ξ𝟏𝟑 𝟑 = 𝑹′ 𝒓 𝟐 𝑾 𝑾 × 𝑾 × 𝟑 ξ𝟏𝟑 𝟐 ξ𝟏𝟑 +𝑹=𝑾→𝑹= = µ × 𝟕𝑾 𝟏𝟑 3 𝑟 2 𝟕𝑾 𝟏𝟑 → µ= 𝟒 𝟕 r ∝ ξ13 r 2 43 9)A metal rod weighing 60gm wt (its weight is not acting at its midpoints), resting at one end on a rough horizontal plane its coefficient of friction is ξ3/2 and its other end rests on a smooth plane inclined at 120 with the first plane. If the rod is just about to slide making a 30 with the horizon find the reactions of the two planes. Determine the ratio at which the point of action of the weight of the rod divides it. The rod AB is in equilibrium under the action of weight 60 at distance x from A , reaction r at A, Limiting friction µ 𝒓 = ξ𝟑 𝟐 𝒓 at A , Reaction at B r’ r cos 30 ξ𝟑 𝟐 𝒓 → 𝒓 = 𝒓′ (1) Moment with respect to A = 0 − 𝟔𝟎 × 𝑨𝑬 + 𝒓′ × 𝑨𝑫 = 𝟎 − 𝟔𝟎 𝒙 𝐜𝐨𝐬 𝟑𝟎𝟎 + 𝒓′ × 𝑳 𝐬𝐢𝐧 𝟔𝟎𝟎 = 𝟎 𝟐 𝟑 30 30 ∑ 𝒀 = 𝟎 → 𝒓′ 𝐬𝐢𝐧 𝟑𝟎𝟎 + 𝒓 = 𝟔𝟎 → 𝒓 = 𝒓′ = 𝟒𝟎 → 𝒙= 30 \ From equilibrium ∑ 𝑿 = 𝟎 → 𝒓′ 𝐜𝐨𝐬 𝟑𝟎𝟎 = r\ sin 30 𝑳 So the weight divides the rod in ratio 2:1 from A 44 The couple ⃑⃑⃑⃑𝟏 = 𝟑𝒊̂ + 𝟒𝒋̂ 𝒂𝒏𝒅 ⃑⃑⃑⃑ 1] 𝑭 𝑭𝟐 𝒇𝒐𝒓𝒎 𝒂 𝒄𝒐𝒖𝒑𝒍𝒆 𝒇𝒊𝒏𝒅 ⃑⃑⃑⃑ 𝑭𝟐 and if F1 acts at A(1,3) and F2 acts at B(3,4) then find the moment of the couple. and the length of the perpendicular distance between the two forces ⃑⃑ ⃑⃑⃑⃑ ⃑⃑⃑⃑𝟐 = 𝑶 𝑭𝟏 + 𝑭 ⃑⃑⃑⃑𝟐 = −𝑭 ⃑⃑⃑⃑𝟏 = −𝟑𝒊̂ − 𝟒𝒋̂ 𝑭 ⃑⃑⃑⃑⃑⃑𝒐 = 𝒐𝑨 ⃑⃑⃑⃑⃑ 𝑿𝑭 ⃑⃑⃑⃑𝟏 + 𝒐𝑩 ⃑⃑⃑⃑⃑ 𝑿𝑭 ⃑⃑⃑⃑𝟐 = 𝑨𝑩 ⃑⃑⃑⃑⃑⃑ 𝑿𝑭 ⃑⃑⃑⃑𝟐 = 𝑩𝑨 ⃑⃑⃑⃑⃑⃑ 𝑿𝑭 ⃑⃑⃑⃑𝟏 = −𝟓𝑲 ⃑⃑⃑ 𝑴 𝑳𝒆𝒏𝒈𝒕𝒉 𝒐𝒇 𝒑𝒆𝒓𝒑𝒆𝒏𝒅𝒊𝒄𝒖𝒍𝒂𝒓 = ‖𝑴‖ 𝟓 = =𝟏 ‖𝑭‖ √(−𝟑)𝟐 + (−𝟒)𝟐 2]Find f if the two couples in equilibrium M1+M2=0 [f][40]+[-6][40]=0 f=6 units 3]ABCD is a parallelogram AB=4cm,BC= 8 cm ,m(A)=120 forces 2N,2N act along ⃑⃑⃑⃑⃑ 𝐁𝐀, ⃑⃑⃑⃑⃑ 𝐃𝐂 and F N,F N act along ⃑⃑⃑⃑⃑ 𝐂𝐁, ⃑⃑⃑⃑⃑⃑ 𝐀𝐃 find F such that the two couples are equivalent. M1=M2 [2][8 sin 60]=[F][4 sin 60] F=4 N 45 4]ABCD is a square of side length 8 cm forces 2N,2N act along ⃑⃑⃑⃑⃑ 𝐀𝐁, ⃑⃑⃑⃑⃑ 𝐂𝐃 and F N,F N act at B and D in directions of ⃑⃑⃑⃑⃑ 𝑪𝑨 𝒂𝒏𝒅 ⃑⃑⃑⃑⃑ 𝑨𝑪 find F such that the two couples are in equilibrium the two diagonals are perpendicular to each other M1+M2=0 (2)(8)+(-f)(8ξ𝟐)=0 so f=ξ𝟐N   5]ABCD is a rectangle, AB=12Cm, BC=5Cm.forces of mag. 10, 10 act a long AB , CD two forces f, f act ⃡⃑⃑⃑⃑ at B, D and parallel to 𝑨𝑪 Find f s.t. the two Couples are in equilibrium. 𝑓 𝑠𝑖𝑛𝜃 F The two forces 10,10 form a Couple The two forces f, f form a couple 𝟓 𝟏𝟐 𝒔𝒊𝒏𝜽 = , 𝒄𝒐𝒔 𝜽 = 𝟏𝟑 𝟏𝟑 𝑴 𝟏 + 𝑴𝟐 = 𝟎 = [𝟏𝟎][𝟓] + [−𝒇 𝒄𝒐𝒔 𝜽][𝟓] + [−𝒇𝒔𝒊𝒏𝜽][𝟏𝟐] = 𝟎 𝒔𝒐 𝒇 = 𝟔𝟓 𝟏𝟐 D A 𝜃 𝑓 𝑐𝑜𝑠𝜃 C 10 𝜃 5 𝜃 12 10 B 𝑓 𝑐𝑜𝑠𝜃 𝜃 F 𝑓 𝑠𝑖𝑛𝜃 𝑵 Another solution We may get the moment by getting it about any point as B 𝑴𝑩 = [𝟏𝟎][𝟓] + [−𝒇 𝐜𝐨𝐬 𝜽][𝟓] + [−𝒇 𝒔𝒊𝒏𝜽][𝟏𝟐] = 𝟎 46 6]A rod whose length 60 cm and weight 10 kg. Wt. acting at its mid point can rotate easily in a vertical plane about a fixed hinge at one end a couple of mag. 150 Kg. Wt cm and whose direction perpendicular to the plane acts on it. Find mag. and direction of the reaction of the hinge and inclination of the rod to the vertical in position of equilibrium The rod is in equilibrium State under the action of a couple and two forces Then these two forces must form a couple r=w and the reaction acts vertically upwards M1+M2=0 150+[-10][30 sin 𝜽]=0 so 𝒔𝒊𝒏 𝜽 = 𝟎. 𝟓 𝒕𝒉𝒆𝒏 𝜽 = 𝟑𝟎 𝒐𝒓 𝟏𝟓𝟎 7] AB is a rod of length 80 Cm and weight 36 N acting at its mid point can rotate easily in a vertical plane about a horizontal pin passing through a hole at a point C 20 Cm from A. If the rod rests with end B on a Smooth horizontal table and end A is Pulled with a rope horizontally till the reaction of the table becomes equal to the weight if the rod inclines by 60 to the horizontal. Find (1) the tension in the rope (2) The reaction of the pin (3)if the string is vertical then the tension= (4)if the string is perpendicular to the rod then the tension= The reaction of the plane and the weight form a couple so the rod is in eq. Under the action of acouple and two forces 47 so these two forces must form a couple M1+M2=0 [36](40 cos 60)+[-T](20 sin 60)=0 𝐓 = 𝟐𝟒ξ𝟑𝐍 [-T][20 cos60]+[36][40 cos 60]=0 then T=72N [36][40 cos60]+[-T][20]=0 then T=36N 48 8]ABCD is a fine lamina in the form of a rectangle AB = 8, BC = 6. Suspended from A and its weight 20N. at the point of intersection of diagonals and kept in eq. by a couple whose moment 100N. cm.Find the inclination of ̅̅̅̅ 𝑨𝑪 to the vertical in position of eq. r M1+M2=0 A 100+[-20][L]=0 6 Then L=5 𝟓 = 𝟏 𝒕𝒉𝒆𝒏 𝜽 = 𝟗𝟎 𝟓 means ̅̅̅̅ 𝑨𝑪 𝒉𝒐𝒓𝒊𝒛𝒐𝒏𝒕𝒂𝒍 8 𝜃 5 L 𝒔𝒊𝒏𝜽 = B X D 100 C 20 9] ABC is a fine lamina in the form of an equilateral triangle, AB = 20, the lamina is suspended freely from A and kept in equilibrium by a couple the magnitude of its moment 1000 N.cm. ̅̅̅̅ if the weight is 200N Find the inclination of 𝑨𝑩 r The lamina is in equilibrium state under the action of a couple and two forces w,r. A So these two forces form a couple. 30 𝜃 WL = 1000  200L = 1000 L = 5cm. x B AXC AX = 20 Sin 60 = 10 C L = 1000 + (−WL) = 0  60 y M = M1 + M 2 In 20 3 200 1000 0 49 2 3 2 3 AY= AX = (10 3 ) = 20 3 3 L 5 15 3 3 = = = = AY 4 20 3 20 3 4 3 ( ) 3 3  = Sin−1 4 Sin = 𝜽 = 𝟐𝟓𝟎 AB inclines to the vertical by angle 300 – 250=50 Take care the angle may equals 550 as we change the order of ABC 10]ABC is a fine lamina in the form of a right angled triangle at B, AB =4, BC = 3, and its weight 10N. Acting at the intersection of its medians if a couple whose direction is perpendicular to the plane of the lamina acts on it so that it is in equilibrium in a position in which AB is vertical. Find the magnitude of the moment of the couple. Given that the triangle is suspended From A. In triangle ABX 𝑨𝒀 𝑳 = 𝑨𝑿 𝑩𝒙 r A L 𝟐 𝑳 = 𝒕𝒉𝒆𝒏 𝑳 = 𝟏 𝟑 𝟏. 𝟓 4 Y X 3 B M1+M2=0 M+(-10)(L) =0 M=10 N.cm M 10 C 50 11] 11 Show that the system form 6 a couple D A C 3 5 B 4 7 10 ⃑ is a unit vector upwards Let 𝑪 ⃑ = ⃑𝟎 it may be in equilibrium or couple ⃑⃑ = [𝟔 + 𝟏𝟏 − 𝟕 − 𝟏𝟎]𝑪 𝑹 𝑴𝑨 = [𝟏𝟏][𝟑] + [−𝟕][𝟖] + [−𝟏𝟎][𝟏𝟐] = −𝟏𝟒𝟑 ≠ 𝟎 the system form a couple 12]ABC is a right angled triangle at B , AB=3 , 𝐁𝐂 = 𝟒 𝐜𝐦, 𝒇𝒐𝒓𝒄𝒆𝒔 𝒐𝒇 𝒎𝒂𝒈𝒏𝒊𝒕𝒖𝒅𝒆𝒔 𝟑𝟎, 𝟒𝟎, 𝟓𝟎𝑵 𝒂𝒄𝒕 𝒂𝒍𝒐𝒏𝒈 ⃑⃑⃑⃑⃑⃑ 𝑨𝑩, ⃑⃑⃑⃑⃑⃑ 𝑩𝑪𝐚𝐧𝐝 ⃑⃑⃑⃑⃑ 𝑪𝑨 𝒓𝒆𝒔𝒑𝒆𝒄𝒕𝒊𝒗𝒆𝒍𝒚 , 𝒑𝒓𝒐𝒗𝒆 𝒕𝒉𝒂𝒕 𝒕𝒉𝒆 𝒔𝒚𝒔𝒕𝒆𝒎 𝒕𝒆𝒏𝒅𝒔 𝒕𝒐 𝒂 𝒄𝒐𝒖𝒑𝒍𝒆 𝒂𝒏𝒅 𝒇𝒊𝒏𝒅 𝒕𝒉𝒆 𝒏𝒐𝒓𝒎 𝒐𝒇 𝒊𝒕𝒔 𝒎𝒐𝒎𝒆𝒏𝒕 A 50 Forces are in the same cyclic order 𝑭 𝟑𝟎 𝑳 𝟑 Scale= = = 𝟒𝟎 𝟒 = 𝟓𝟎 𝟓 = 𝟏𝟎 M=𝟐 × 𝒂𝒓𝒆𝒂 × 𝒔𝒄𝒂𝒍𝒆 𝟏 = 𝟐 × [ × 𝟒 × 𝟑] × 𝟏𝟎 = 𝟏𝟐𝟎 𝑵. 𝒄𝒎 𝟐 Another solution 5 3 B 30 40 4 C 𝑴𝑨 = [𝟒𝟎][3]=120 N.cm 𝑴𝑩 = [𝟓𝟎] [ 𝟑×𝟒 𝟓 ]=120 N.cm 𝑴𝑪 = [𝟑𝟎][𝟒]=120 N.cm 𝑴𝑨 = 𝑴𝑩 = 𝑴𝑪 ≠ 𝟎 𝒕𝒉𝒆 𝒔𝒚𝒔𝒕𝒆𝒎 𝒇𝒐𝒓𝒎 𝒂 𝒄𝒐𝒖𝒑𝒍𝒆 𝒘𝒉𝒐𝒔𝒆 𝒎𝒐𝒎𝒆𝒏𝒕 𝟏𝟐𝟎𝑵. 𝒄𝒎 51 ̅̅̅̅, AB= 15 cm , BC=7.5 cm,CD=9 cm, ̅̅̅̅ 𝒑𝒂𝒓𝒂𝒍𝒍𝒆𝒍 𝒕𝒐 𝑫𝑪 13]ABCD is a trapezium ,m(A)=90,𝑨𝑩 ⃑⃑⃑⃑⃑⃑ , ⃑⃑⃑⃑⃑⃑ 𝒇𝒐𝒓𝒄𝒆𝒔 𝒂𝒄𝒕 𝒂𝒍𝒐𝒏𝒈 ⃑⃑⃑⃑⃑⃑ 𝑨𝑩, ⃑⃑⃑⃑⃑⃑ 𝑩𝑪,𝑪𝑫 𝑫𝑨 𝒓𝒆𝒔𝒑𝒆𝒄𝒕𝒊𝒗𝒆𝒍𝒚 , 𝒊𝒇 𝒕𝒉𝒆 𝒔𝒚𝒔𝒕𝒆𝒎 𝒕𝒆𝒏𝒅𝒔 𝒕𝒐 𝒂 𝒄𝒐𝒖𝒑𝒍𝒆 𝒘𝒉𝒐𝒔𝒆 𝒎𝒐𝒎𝒆𝒏𝒕 𝒆𝒒𝒖𝒂𝒍𝒔 𝟒𝟑𝟐 𝒈𝒎. 𝒘𝒕. 𝒄𝒎 In the direction of ABCD ,find the magnitude of each force D 𝟏 𝒂𝒓𝒆𝒂 = [𝟗 × 𝟒. 𝟓 + × 𝟔 × 𝟒. 𝟓] = 𝟓𝟒 𝟐 f2 4.5 f4 𝑴 = 𝟐 × 𝒂𝒓𝒆𝒂 × 𝒔𝒄𝒂𝒍𝒆 9 𝟒𝟑𝟐 = 𝟐 × [𝟓𝟒]𝒔𝒄𝒂𝒍𝒆 𝟒𝟑𝟐 =𝟒 𝟏𝟎𝟖 7.5 6 f1 A 𝒔𝒄𝒂𝒍𝒆 = 𝒔𝒄𝒂𝒍𝒆 = C f3 B 15 𝑭𝟏 𝑭𝟐 𝑭𝟑 𝑭𝟒 = = = =𝟒 𝟏𝟓 𝟕. 𝟓 𝟗 𝟒. 𝟓 𝑭𝟏 = 𝟔𝟎 𝒈𝒎. 𝒘𝒕 , 𝑭𝟐 = 𝟑𝟎𝒈𝒎. 𝒘𝒕, 𝑭𝟑 = 𝟑𝟔𝒈𝒎. 𝒘𝒕 , 𝑭𝟒 = 𝟏𝟖𝒈𝒎. 𝒘𝒕 14]ABCDE is a regular pentagon of side length 10 cm ,forces 5 N act along the sides in the same cyclic order show that the system form a couple and get the norm of the moment forces are in the same cyclic order 𝒔𝒄𝒂𝒍𝒆 = 𝑭 𝑳 = 𝟓 𝟏𝟎 = 5 5 𝟏 𝟐 5 5 5 36 𝒏 𝟏𝟖𝟎 𝟒 𝒏 Area of a regular polygon = 𝑺𝟐 𝒄𝒐𝒕 𝒏 𝟏𝟖𝟎 𝟒 𝒏 Area of a regular pentagon = 𝑺𝟐 𝒄𝒐𝒕 𝟓 𝟏𝟖𝟎 𝟒 𝟓 = [𝟏𝟎]𝟐 𝒄𝒐𝒕 𝑴𝒐𝒎𝒆𝒏𝒕 = 𝟐𝒂𝒓𝒆𝒂. 𝒔𝒄𝒂𝒍𝒆 = 𝟐[𝟏𝟕𝟐. 𝟎𝟒𝟖] × =𝟏𝟐𝟓 × 𝟏 = 𝟏𝟕𝟐. 𝟎𝟒𝟖 𝑵. 𝒄𝒎 𝟐 36 𝟏 𝒕𝒂𝒏 𝟑𝟔 = 𝟏𝟕𝟐. 𝟎𝟒𝟖 𝒄𝒎𝟐 52 ̅̅̅̅ , 15]ABCD is a rectangle in which AB=30cm, BC=40 cm , O is the mid point of 𝑪𝑫 ̅̅̅̅, 𝒇𝒐𝒓𝒄𝒆𝒔 𝒐𝒇 𝒎𝒂𝒈𝒏𝒊𝒕𝒖𝒅𝒆𝒔 𝟒𝟎, 𝟔𝟎, 𝟒𝟎, 𝟔𝟎, 𝟓𝟎, 𝟓𝟎 𝑵 𝑯 𝒊𝒔 𝒕𝒉𝒆 𝒎𝒊𝒅 𝒑𝒐𝒊𝒏𝒕 𝒐𝒇 𝑨𝑫 ⃑⃑⃑⃑⃑⃑ , 𝑫𝑪 ⃑⃑⃑⃑⃑⃑ , 𝑨𝑫 ⃑⃑⃑⃑⃑ , 𝑶𝑯 ⃑⃑⃑⃑⃑⃑ 𝒓𝒆𝒔𝒑𝒆𝒄𝒕𝒊𝒗𝒆𝒍𝒚 , 𝒇𝒊𝒏𝒅 𝒕𝒉𝒆 𝒏𝒐𝒓𝒎 𝒐𝒇 𝒕𝒉𝒆 𝒓𝒆𝒔𝒖𝒍𝒕𝒂𝒏𝒕 𝒄𝒐𝒖𝒑𝒍𝒆 ⃑⃑⃑⃑⃑⃑ , 𝑪𝑩 ⃑⃑⃑⃑⃑⃑ , 𝑨𝑪 𝒂𝒄𝒕 𝒂𝒍𝒐𝒏𝒈 𝑩𝑨 The two forces 40N and 40 N act along ⃑⃑⃑⃑⃑⃑ 𝑩𝑨 𝒂𝒏𝒅 ⃑⃑⃑⃑⃑⃑ 𝑫𝑪 𝒇𝒐𝒓𝒎 𝒂 𝒄𝒐𝒖𝒑𝒍𝒆 The two forces 60N and 60 N act along ⃑⃑⃑⃑⃑⃑ 𝑨𝑫 𝒂𝒏𝒅 ⃑⃑⃑⃑⃑⃑ 𝑪𝑩 𝒇𝒐𝒓𝒎 𝒂 𝒄𝒐𝒖𝒑𝒍𝒆 The two forces 50N and 50 N act along ⃑⃑⃑⃑⃑ 𝒂𝒏𝒅 𝑶𝑯 ⃑⃑⃑⃑⃑⃑ 𝒇𝒐𝒓𝒎 𝒂 𝒄𝒐𝒖𝒑𝒍𝒆 𝑨𝑪 The system form a couple 𝑴 = [−𝟒𝟎][𝟒𝟎] + [−𝟔𝟎][𝟑𝟎] + [𝟓𝟎 𝐬𝐢𝐧𝛉][𝟐𝟎] 𝑴 = −𝟐𝟖𝟎𝟎 𝐍. 𝐜𝐦 𝒕𝒉𝒆𝒏 𝒏𝒐𝒓𝒎 𝒐𝒇 𝒎𝒐𝒎𝒆𝒏𝒕 = 𝟐𝟖𝟎𝟎 𝑵. 𝒄𝒎 16]ABCD is a rectangle AB=8 cm , BC=15 cm, 𝒇𝒐𝒓𝒄𝒆𝒔 𝒐𝒇 𝒎𝒂𝒈𝒏𝒊𝒕𝒖𝒅𝒆𝒔 𝟏𝟗, 𝟒𝟓, ⃑⃑⃑⃑⃑⃑ ,𝑪𝑫 ⃑⃑⃑⃑⃑⃑ , 𝑫𝑨 ⃑⃑⃑⃑⃑ 𝒓𝒆𝒔𝒑𝒆𝒄𝒕𝒊𝒗𝒆𝒍𝒚 ,Find the algebraic component of the ⃑⃑⃑⃑⃑⃑ , 𝑩𝑪 ⃑⃑⃑⃑⃑⃑ , 𝑪𝑨 𝟑, 𝟏𝟓, 𝟑𝟒 𝑵 𝒂𝒄𝒕 𝒂𝒍𝒐𝒏𝒈 𝑨𝑩 force 34 in the direction of ⃑⃑⃑⃑⃑⃑ 𝑪𝑩 𝒂𝒏𝒅 ⃑⃑⃑⃑⃑⃑ 𝑪𝑫 𝒕𝒉𝒆𝒏 𝒑𝒓𝒐𝒗𝒆 𝒕𝒉𝒂𝒕 𝒕𝒉𝒆 𝒔𝒚𝒔𝒕𝒆𝒎 𝒕𝒆𝒏𝒅𝒔 𝒕𝒐 𝒂 𝒄𝒐𝒖𝒑𝒍𝒆 𝒂𝒏𝒅 𝒇𝒊𝒏𝒅 𝒕𝒉𝒆 𝒏𝒐𝒓𝒎 𝒐𝒇 𝒊𝒕𝒔 𝒎𝒐𝒎𝒆𝒏𝒕 M=(19)(15)+(15)(8)=405 N.cm 53 Another solution We may get the moment about three non collinear points if they are equal to zero then its in equilibrium but if they equal but not to zero then it’s a couple 𝐌𝐀 = [𝟒𝟓][𝟖] + [𝟑][𝟏𝟓] = 𝟒𝟎𝟓 𝐍. 𝐜𝐦 𝐌𝐂 = [𝟏𝟗][𝟏𝟓] + [𝟏𝟓][𝟖] = 𝟒𝟎𝟓 𝐍. 𝐜𝐦 𝐌𝐁 = [𝟏𝟓][𝟖] + [𝟑][𝟏𝟓] + [𝟑𝟒][ 𝟖 × 𝟏𝟓 ] = 𝟒𝟎𝟓 𝐍. 𝐜𝐦 𝟏𝟕 𝐌𝐀 = 𝐌𝐁 = 𝐌𝐜 = 𝟒𝟎𝟓 𝑵. 𝒄𝒎 𝒊𝒕𝒔 𝒂 𝒄𝒐𝒖𝒑𝒍𝒆 𝒘𝒉𝒐𝒔𝒆 𝒎𝒐𝒎𝒆𝒏𝒕 𝟒𝟎𝟓 𝑵. 𝒄𝒎 54 Center of gravity 1]Two bodies of masses 6kg and 12 kg are suspended from the ends of a light horizontal rod of length 90 cm find the center of gravity of the two weights . A B W 6 12 X 0 90 Y 0 0 𝒙𝑮 = B(90,0) 6 12 𝒙𝟏 𝒘𝟏 + 𝒙𝟐 𝒘𝟐 [𝟎][𝟔] + [𝟗𝟎][𝟏𝟐] = = 𝟔𝟎 𝒘𝟏 + 𝒘𝟐 𝟔 + 𝟏𝟐 𝒚𝑮 = its 60 cm from A A(0,0) 𝒚𝟏 𝒘𝟏 + 𝒚𝟐 𝒘𝟐 [𝟎][𝟔] + [𝟎][𝟏𝟐] = =𝟎 𝒘𝟏 + 𝒘𝟐 𝟔 + 𝟏𝟐 G=(60,0) 2]Find the center of gravity of the following distribution𝑭𝟏 = 𝟑𝑵 𝒂𝒕 (𝟒, −𝟏) 𝑭𝟐 = 𝟓𝑵 𝒂𝒕 (𝟎, 𝟑), 𝑭𝟑 = 𝟒𝑵 𝒂𝒕 (−𝟐, 𝟑) and the forces are like[having the same direction] A B C W 3 5 4 X 4 0 -2 Y -1 3 3 𝒙𝑮 = 𝒚𝑮 = [−𝟏][𝟑]+[𝟑][𝟓]+[𝟑][𝟒] 𝟑+𝟓+𝟒 =𝟐 [𝟒][𝟑] + [𝟎][𝟓] + [−𝟐][𝟒] 𝟏 = 𝟑+𝟓+𝟒 𝟑 𝟏 𝑮 = ( ,2) 𝟑 55 3]A wire of uniform thickness and density in the form of trapezium ABCD,AB=15cm,BC=12cm,CD=10 cm,m(ABC)=m(BCD)=900 ̅̅̅̅ ̅̅̅̅, 𝑩𝑪 Find the distance between the center of gravity and the sides 𝑨𝑩 C D 10 12 12 5 B A 10 A(15,0),B(0,0),C(0,12), D(10,12) ̅̅̅̅ is 15 m acts at E its mid point The weight of 𝑨𝑩 ̅̅̅̅ is 12 m acts at F its mid point The weight of 𝑩𝑪 ̅̅̅̅ is 10 m acts at G its mid point The weight of 𝑪𝑫 The weight of ̅̅̅̅ 𝑨𝑫 is 13 m acts at H its mid point 𝒙𝑮 = 𝒚𝑮 = E F G H W 15m 12m 10m 13m X 7.5 0 5 12.5 Y 0 6 12 6 [𝟕. 𝟓][𝟏𝟓𝒎] + [𝟎][𝟏𝟐𝒎] + [𝟓][𝟏𝟎𝒎] + [𝟏𝟐. 𝟓][𝟏𝟑𝒎] = 𝟔. 𝟓 𝟏𝟓𝒎 + 𝟏𝟐𝒎 + 𝟏𝟎𝒎 + 𝟏𝟑𝒎 [𝟎][𝟏𝟓𝒎]+[𝟔][𝟏𝟐𝒎]+[𝟏𝟐][𝟏𝟎𝒎]+[𝟔][𝟏𝟑𝒎] 𝟏𝟓𝒎+𝟏𝟐𝒎+𝟏𝟎𝒎+𝟏𝟑𝒎 = 𝟓. 𝟒 G=(6.5,5.4) ̅̅̅̅=𝒚𝑮 = 𝟓. 𝟒 𝒄𝒎 The distance from 𝑨𝑩 ̅̅̅̅=𝒙𝑮 = 𝟔. 𝟓 𝒄𝒎 The distance from 𝑩𝑪 Another solution We may divide the weight equally at the ends of the line segment ̅̅̅̅ is 15 m acts as 7.5 m at A and 7.5 m at B The weight of 𝑨𝑩 ̅̅̅̅ is 12 m acts as 6 m at B and 6 m at C The weight of 𝑩𝑪 ̅̅̅̅ is 10 m acts as 5 m at C and 5 m at D The weight of 𝑪𝑫 56 The weight of ̅̅̅̅ 𝑨𝑫 is 13 m acts as 6.5 m at A and 6.5 m at D The total weight at A is 14m The total weight at B is 13.5 m The total weight at C is 11m The total weight at D is 11.5m 𝒙𝑮 = 𝒚𝑮 = A B C D W 14m 13.5m 11m 11.5m X 15 0 0 10 Y 0 0 12 12 [𝟏𝟓][𝟏𝟒𝒎] + [𝟎][𝟏𝟑. 𝟓𝒎] + [𝟎][𝟏𝟏𝒎] + [𝟏𝟎][𝟏𝟏. 𝟓𝒎] = 𝟔. 𝟓 𝟏𝟒𝒎 + 𝟏𝟑. 𝟓𝒎 + 𝟏𝟏𝒎 + 𝟏𝟏. 𝟓𝒎 [𝟎][𝟏𝟒𝒎]+[𝟎][𝟏𝟑.𝟓𝒎]+[𝟏𝟐][𝟏𝟏𝒎]+[𝟏𝟐][𝟏𝟏.𝟓𝒎] 𝟏𝟒𝒎+𝟏𝟑.𝟓𝒎+𝟏𝟏𝒎+𝟏𝟏.𝟓𝒎 = 𝟓. 𝟒 G(6.5,5.4) 4]A uniform wire of length 100 cm,is bent in the form of five sides of regular hexagon ABCDEF, find the distance between the center of the hexagon and the center of gravity 1 the length of each part=100/5=20 B(− 2 𝐿, ξ3 𝐿) 2 A(10,10ξ3) the weight of each part=20m C(-𝐿, 0) 1 D(− 2 𝐿, − hexagon -line W 120m -20m X 0 𝟏𝟓 Y 0 𝟓ξ𝟑 𝒙𝑮 = F(20,0) ξ3 𝐿) 2 1 E(2 𝐿, − ξ3 𝐿) 2 [𝟎][𝟏𝟐𝟎𝒎] − [𝟑𝟎𝟎𝒎] [𝟎][𝟏𝟐𝟎𝒎] − [𝟏𝟎𝟎ξ𝟑𝒎] = −𝟑 , 𝒚𝑮 = = −ξ𝟑 𝟏𝟎𝟎𝒎 𝟏𝟎𝟎𝒎 𝟐 distance=√(−𝟑)𝟐 + (−ξ𝟑) = 𝟐ξ𝟑 𝒄𝒎 57 5]A uniform squared lamina of weight W is suspended freely from the vertex A and a weight of 𝟏 𝟒 ̅̅̅̅ to the vertical 𝒘 is fixed at B, prove that the tangent of the angle of inclination of the diagonal 𝑨𝑪 in equilibrium is 𝟏 𝟓 , B(0,0) , C(L,0) ,A(0,L),D(L,L) the original center of gravity is the point of intersection of diagonals after adding the weight at B A let’s get the new center of gravity G1 B W W 0.25W X 0.5L 0 Y 0.5L 0 D 𝜃 G B C [𝟎. 𝟓𝑳]𝒘 + [𝟎][𝟎. 𝟐𝟓𝒘] = 𝟎. 𝟒𝑳 𝒘 + 𝟎. 𝟐𝟓𝒘 [𝟎. 𝟓𝑳]𝒘 + [𝟎][𝟎. 𝟐𝟓𝒘] 𝒚𝑮 = = 𝟎. 𝟒𝑳 𝒘 + 𝟎. 𝟐𝟓𝒘 𝒙𝑮 = ⃑⃑⃑⃑⃑ . ⃑⃑⃑⃑⃑ |(𝟎. 𝟓𝑳, −𝟎. 𝟓𝑳). (𝟎. 𝟒𝑳, −𝟎. 𝟔𝑳)|. 𝑨𝑮| 𝟓 |𝑨𝑪 𝐜𝐨𝐬 𝜽 = = = ⃑⃑⃑⃑⃑ ‖‖𝑨𝑮 ⃑⃑⃑⃑⃑ ‖ ξ𝟎. 𝟐𝟓𝑳𝟐 + 𝟎. 𝟐𝟓𝑳𝟐 ξ𝟎. 𝟏𝟔𝑳𝟐 + 𝟎. 𝟑𝟔𝑳𝟐 ξ𝟐𝟔 ‖𝑨𝑪 𝒕𝒉𝒆𝒏 𝐭𝐚𝐧 𝜽 = 𝟏 𝟓 6]ABCD is a fine lamina of uniform density in the form of a square of side length 8 cm, A circular ̅̅̅̅ and 3 cm from ̅̅̅̅ part of radius 2 cm and its center is 3 cm from 𝑨𝑩 𝑨𝑫 is removed then suspended ̅̅̅̅ ,the midpoint of ̅̅̅̅ at B,and a triangle is fixed such that its vertices at C,the midpoint of 𝑩𝑪 𝑪𝑫 and the density of its material is twice the density of the material of the square. Find the new center of gravity ̅̅̅̅ to the vertical. If the lamina is suspended from A find the inclination of 𝑨𝑩 58 there is a ratio between area and weight F(4,8) D(0,8) C(8,8) area of square=64 area of circle= 𝝅(𝟐𝟐 ) = 𝟖𝟖 𝟕 𝟏 area of triangle= × 𝟒 × 𝟒 = 𝟖 𝟐 G E(8,4) A(0,0) B(8,0) 𝜃 Vertical the density of the triangle is twice the density of the square Square:circle:triangle 64 : = 8 : 𝟖𝟖 𝟕 𝟏𝟏 𝟕 = 56: 11 :16 :2 divide by 8 multiply by 7 :14 the weight of the square=56m the weight of the circle=11m the weight of the triangle=14m note that we may use [64m, 𝟖𝟖 𝟕 m,16 m] and get the same result 𝟖+𝟖+𝟒 𝟒+𝟖+𝟖 the centroid of the triangle FCE=( 𝟑 , 𝟑 𝟐𝟎 𝟐𝟎 )= (𝟑 , 𝟑 ) square circle circle triangle W 56m -11m 11m 14m X 4 3 8 Y 4 3 0 𝟐𝟎 𝟑 𝟐𝟎 𝟑 𝟐𝟎 ] [𝟏𝟒𝒎] 𝟏𝟏𝟏𝟕 𝟑 𝒙𝑮 = = = 𝟓. 𝟑𝟏𝟗 𝟓𝟔𝒎 − 𝟏𝟏𝒎 + 𝟏𝟏𝒎 + 𝟏𝟒𝒎 𝟐𝟏𝟎 𝟐𝟎 [𝟒][𝟓𝟔𝒎] + [𝟑][−𝟏𝟏𝒎] + [𝟎][𝟏𝟏𝒎] + [ ] [𝟏𝟒𝒎] 𝟖𝟓𝟑 𝟑 𝒀𝑮 = = = 𝟒. 𝟎𝟔𝟏𝟗 𝟓𝟔𝒎 − 𝟏𝟏𝒎 + 𝟏𝟏𝒎 + 𝟏𝟒𝒎 𝟐𝟏𝟎 [𝟒][𝟓𝟔𝒎] + [𝟑][−𝟏𝟏𝒎] + [𝟖][𝟏𝟏𝒎] + [ 59 ⃑⃑⃑⃑⃑ . 𝑨𝑩 ⃑⃑⃑⃑⃑⃑ | |𝑨𝑮 𝒄𝒐𝒔 𝜽 = = ⃑⃑⃑⃑⃑ ‖. ‖𝑨𝑩 ⃑⃑⃑⃑⃑⃑ ‖ ‖𝑨𝑮 = 𝟏𝟏𝟏𝟕 𝟖𝟓𝟑 |( , . (𝟖, 𝟎)| 𝟐𝟏𝟎 𝟐𝟏𝟎) 𝟐 𝟐 √(𝟏𝟏𝟏𝟕) + (𝟖𝟓𝟑) √(𝟖)𝟐 + (𝟎)𝟐 𝟐𝟏𝟎 𝟐𝟏𝟎 𝟏𝟏𝟏𝟕 𝟖𝟓𝟑 |( , . (𝟖, 𝟎)| 𝟐𝟏𝟎 𝟐𝟏𝟎) = 𝟎. 𝟕𝟗𝟒𝟕𝟔 𝟐 𝟐 √(𝟏𝟏𝟏𝟕) + (𝟖𝟓𝟑) √(𝟖)𝟐 + (𝟎)𝟐 𝟐𝟏𝟎 𝟐𝟏𝟎 𝜽 = 𝟑𝟕. 𝟑𝟔𝟕𝟐 7]A uniform fine circular board whose center is the origin and radius is 24 cm, two circular discs whose center the center of one of them is (-2,-12) and radius length 4 cm, where the center of the other is (6,10) and radius length 12 cm are cut off find the center of gravity of the remaining part . Area of board = 𝝅(𝟐𝟒𝟐 ) = 𝟓𝟕𝟔𝝅 Area of circle1= 𝝅(𝟒𝟐 ) = 𝟏𝟔𝝅 Area of circle2= 𝝅(𝟏𝟐𝟐 ) = 𝟏𝟒𝟒𝝅 Weight of board:weight of circle1:weight of circle2 𝟓𝟕𝟔𝝅 : 𝟏𝟔𝝅 : 𝟏𝟒𝟒𝝅 36 : 𝟏 : 9 weight of board=36m , weight of circle1=m , weight of circle2=9m board Circle1 Circle2 W 36m -m -9m X 0 -2 6 Y 0 -12 10 𝒙𝑮 = 𝒚𝑮 = G(-2,-3) [𝟎][𝟑𝟔𝒎] + [−𝟐][−𝒎] + [𝟔][−𝟗𝒎] = −𝟐 𝟐𝟔𝒎 [𝟎][𝟑𝟔𝒎] + [−𝟏𝟐][−𝒎] + [𝟏𝟎][−𝟗𝒎] = −𝟑 𝟐𝟔𝒎 60 8]A fine lamina of uniform density in the form of rectangle ABCD,AB=25 cm,BC=16cm,E on ̅̅̅̅,BE=10cm, F on 𝑨𝑩 ̅̅̅̅ ̅̅̅̅ triangle BEF is cut off and the lamina rests in a vertical plane such that 𝑪𝑬 𝑩𝑪 coincide with a smooth horizontal table, then the lamina is about to rotate about E find BF The area of rectangle =𝟐𝟓 × 𝟏𝟔 = 𝟒𝟎𝟎 D(0,25) 𝟏 The area of triangle = × 𝟏𝟎 × 𝒚 = 𝟓𝒚 A(16,25) 𝟐 F(16,y) 𝒂𝒓𝒆𝒂 𝒐𝒇 𝒙𝒚𝒛 𝟓𝒚 𝒚 = = 𝒂𝒓𝒆𝒂 𝒐𝒇 𝑨𝑩𝑪 𝟒𝟎𝟎 𝟖𝟎 weight of rectangle=80m weight of triangle =ym C(0,0) rectangle triangle W 80m -ym X 8 Y 12.5 E(6,0) B(16,0) 𝟑𝟖 𝟑 𝒚 𝟑 As it about to overturn the center of gravity vertically above E so its x coordinate is 6 𝒙𝑮 = 𝟔𝟒𝟎 − 𝟑𝟖 𝟑 𝒚 = 𝟒𝟖𝟎 − 𝟔𝒚 then 𝟐𝟎 𝟑 𝟑𝟖 ] [−𝒚𝒎] 𝟑 =𝟔 [𝟖𝟎 − 𝒚]𝒎 [𝟖][𝟖𝟎𝒎] + [ 𝒚 = 𝟏𝟔𝟎 𝒔𝒐 𝒚 = 𝟐𝟒 𝒄𝒎

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