Introduction • This chapter builds on chapter 3 and focuses on objects in equilibrium, ie) On the point of moving but actually remaining stationary • As in chapter 3 it involves resolving forces in different directions • Statics is important in engineering for calculating whether structures are stable Statics of a Particle y You can solve problems involving particles in equilibrium by considering forces acting horizontally and vertically 4N 4Sin45 Similar to chapter 3, for these types of problem you should: 1) Draw a diagram and label the forces 45° 4Cos45 30° PCos30 The particle to the PN left is in equilibrium. Calculate the magnitude of the PSin30 forces P and Q. x 2) Resolve into horizontal/vertical or parallel/perpendicular components 3) Set the sums equal to 0 (as the objects are in equilibrium, the forces acting in opposite directions must cancel out… 4) Solve the equations to find the unknown forces… ο This means the horizontal and vertical forces cancel out (acceleration = 0 in both directions so F = 0) QN Resolve Horizontally πΉ = ππ ππΆππ 30 − 4πΆππ 45 = 0 Choose a direction as positive and sub in values Rearrange ππΆππ 30 = 4πππ 45 4πππ 45 πΆππ 30 π = 3.27π π= Divide by Cos30 Calculate 4A Statics of a Particle y You can solve problems involving particles in equilibrium by considering forces acting horizontally and vertically 4N 4Sin45 Similar to chapter 3, for these types of problem you should: 1) Draw a diagram and label the forces 45° 4Cos45 30° PCos30 The particle to the PN left is in equilibrium. Calculate the magnitude of the PSin30 forces P and Q. x 2) Resolve into horizontal/vertical or parallel/perpendicular components 3) Set the sums equal to 0 (as the objects are in equilibrium, the forces acting in opposite directions must cancel out… 4) Solve the equations to find the unknown forces… ο This means the horizontal and vertical forces cancel out (acceleration = 0 in both directions so F = 0) P = 3.27N QN Resolve Vertically πΉ = ππ ππ ππ30 + 4πππ45 − π = 0 Choose a direction as positive and sub in values Add Q ππ ππ30 + 4πππ45 = π 4.46 = π Calculate Q using the exact value of P from the first part You will usually need to identify which direction is solvable first, then solve the second direction after! 4A Statics of a Particle y You can solve problems involving particles in equilibrium by considering forces acting horizontally and vertically The diagram to the right shows a particle in equilibrium under a number of forces. QN PN 1N QSin55 PSin40 55° QCos55 Calculate the magnitudes of the forces P and Q ο Start by resolving in both directions 1) ππΆππ 40 − ππΆππ 55 = 0 2) ππππ40 + ππππ55 − 1 = 0 40° PCos40 x 2N Resolve Horizontally πΉ = ππ ππΆππ 40 − ππΆππ 55 = 0 Choose a direction as positive and sub in values Resolve Vertically πΉ = ππ ππππ40 + ππππ55 + 1 − 2 = 0 ππππ40 + ππππ55 − 1 = 0 Choose a direction as positive and sub in values Simplify 4A Statics of a Particle y QN You can solve problems involving particles in equilibrium by considering forces acting horizontally and vertically PN 1N QSin55 PSin40 55° The diagram to the right shows a particle in equilibrium under a number of forces. QCos55 Calculate the magnitudes of the forces P and Q 2) ππππ40 + ππππ55 − 1 = 0 x PCos40 2N ο Start by resolving in both directions 1) ππΆππ 40 − ππΆππ 55 = 0 40° 2) ππΆππ 55 π= πΆππ 40 ππππ40 + ππππ55 − 1 = 0 ππΆππ 55 πππ40 + ππππ55 − 1 = 0 πΆππ 40 Replace P with the Q equivalent Multiply all terms by Cos40 ππΆππ 55πππ40 + ππππ55πΆππ 40 − πΆππ 40 = 0 ο You can now solve these by rearranging one and subbing it into the other! Divide by ο Q = 0.769N the bracket Add Cos40 ππΆππ 55πππ40 + ππππ55πΆππ 40 = πΆππ 40 π(πΆππ 55πππ40 + πππ55πΆππ 40) = πΆππ 40 π= Calculate Factorise Q on the left side πΆππ 40 (πΆππ 55πππ40 + πππ55πΆππ 40) π = 0.769π 4A Statics of a Particle y QN You can solve problems involving particles in equilibrium by considering forces acting horizontally and vertically PN 1N QSin55 PSin40 55° The diagram to the right shows a particle in equilibrium under a number of forces. QCos55 Calculate the magnitudes of the forces P and Q 40° PCos40 x 2N ο Start by resolving in both directions 1) ππΆππ 40 − ππΆππ 55 = 0 π= ππΆππ 55 πΆππ 40 1) π= 2) ππππ40 + ππππ55 − 1 = 0 ο You can now solve these by rearranging one and subbing it into the other! ο Q = 0.769N ο P = 0.576N π= ππΆππ 55 πΆππ 40 0.769πΆππ 55 πΆππ 40 Sub in Q (use the exact value) Calculate π = 0.576π 4A Statics of a Particle PN You can solve problems involving particles in equilibrium by considering forces acting horizontally and vertically θ The diagram shows a particle in equilibrium on an inclined plane under the effect of the forces shown. Find the magnitude of the force P and the size of angle θ. ο Start by splitting forces into parallel and perpendicular directions 1) ππΆππ π = 5πππ30 + 8 2) πππππ = 5πΆππ 30 − 2 PSinθ 2N PCosθ 5Cos30 30° 5N 8N 5Sin30 30° Resolving Parallel πΉ = ππ ππΆππ π − 5πππ30 − 8 = 0 Use P as the positive direction and sub in values ππΆππ π = 5πππ30 + 8 Rearrange to leave PCosθ Resolving Perpendicular πΉ = ππ πππππ + 2 − 5πΆππ 30 = 0 Use P as the positive direction and sub in values πππππ = 5πΆππ 30 − 2 Rearrange to leave PSinθ 4A Statics of a Particle PN You can solve problems involving particles in equilibrium by considering forces acting horizontally and vertically θ The diagram shows a particle in equilibrium on an inclined plane under the effect of the forces shown. Find the magnitude of the force P and the size of angle θ. ο Start by splitting forces into parallel and perpendicular directions 1) ππΆππ π = 5πππ30 + 8 2) πππππ = 5πΆππ 30 − 2 π = 12.5° PSinθ 2N PCosθ 5Cos30 30° 5N 8N 30° 2) πππππ = 5πΆππ 30 − 2 1) ππΆππ π = 5πππ30 + 8 ππππ = 5πΆππ 30 − 2 5πππ30 + 8 ππππ = 0.2219 … π = 12.5° 5Sin30 Divide equation 2 by equation 1 ο Each side must be divided as a whole, not individual parts ο P’s cancel, Sin/Cos = Tan Work out the fraction Use inverse Tan 4A Statics of a Particle PN You can solve problems involving particles in equilibrium by considering forces acting horizontally and vertically θ The diagram shows a particle in equilibrium on an inclined plane under the effect of the forces shown. Find the magnitude of the force P and the size of angle θ. ο Start by splitting forces into parallel and perpendicular directions 1) ππΆππ π = 5πππ30 + 8 2) πππππ = 5πΆππ 30 − 2 PCosθ 5Cos30 30° 5N 8N 30° 1) ππΆππ π = 5πππ30 + 8 5πππ30 + 8 π= πΆππ π π= 5πππ30 + 8 πΆππ 12.5 π = 12.5° π = 10.8π PSinθ 2N 5Sin30 Divide by Cosθ Sub in the exact value for θ Calculate P π = 10.8π 4A Statics of a Particle Q You need to know when to include additional forces on your diagrams, such as weight, tension, thrust, the normal reaction and friction A particle of mass 3kg is held in equilibrium by two light inextensible strings. One of the strings is horizontal, and the other is inclined at 45° to the horizontal, as shown. The tension in the horizontal string is P and in the other string is Q. Find the values of P and Q. π = 41.6π QSin45 45° P QCos45 3g Resolve vertically πΉ = ππ ππππ45 − 3π = 0 ππππ45 = 3π 3π π= πππ45 Choosing Q as the positive direction, sub in values… Add 3g Divide by Sin45 Calculate π = 41.6π 4B Statics of a Particle Q You need to know when to include additional forces on your diagrams, such as weight, tension, thrust, the normal reaction and friction A particle of mass 3kg is held in equilibrium by two light inextensible strings. One of the strings is horizontal, and the other is inclined at 45° to the horizontal, as shown. The tension in the horizontal string is P and in the other string is Q. Find the values of P and Q. QSin45 45° P QCos45 3g Resolve horizontally πΉ = ππ π = 41.6π ππΆππ 45 − π = 0 π = 29.4π ππΆππ 45 = π 41.6πΆππ 45 = π Choosing Q as the positive direction, sub in values… Add P Sub in the value of Q from before Calculate P 29.4 = π 4B Statics of a Particle X You need to know when to include additional forces on your diagrams, such as weight, tension, thrust, the normal reaction and friction A smooth bead, Y, is threaded on a light inextensible string. The ends of the string are attached to two fixed points X and Z on the same horizontal level. The bead is held in equilibrium by a horizontal force of 8N acting in the direction ZX. Bead Y hangs vertically below X and angle XZY = 30°. Z 30° T T 8 Y TCos30 π = 9.24π ο Since this is only one string and it is inextensible, the tension in it will be the same ο Call the mass m, since we do not know it… mg Resolve Horizontally πΉ = ππ ππΆππ 30 − 8 = 0 Find: a) The tension in the string b) The weight of the bead TSin30 30° Draw a diagram ππΆππ 30 = 8 8 π= πΆππ 30 Sub in values, choosing T as the positive direction Add 8 Divide by Cos30 Calculate π = 9.24π 4B Statics of a Particle X You need to know when to include additional forces on your diagrams, such as weight, tension, thrust, the normal reaction and friction A smooth bead, Y, is threaded on a light inextensible string. The ends of the string are attached to two fixed points X and Z on the same horizontal level. The bead is held in equilibrium by a horizontal force of 8N acting in the direction ZX. Bead Y hangs vertically below X and angle XZY = 30°. Z 30° T T 8 Y 30° TCos30 ο Since this is only one string and it is inextensible, the tension in it will be the same ο Call the mass m, since we do not know it… mg Resolve Vertically πΉ = ππ ππππ30 + π − ππ = 0 Find: a) The tension in the string b) The weight of the bead TSin30 Draw a diagram ππππ30 + π = ππ 9.24πππ30 + 9.24 = ππ π = 9.24π Sub in values, choosing T as the positive direction Add mg Sub in the value of T This is all we need! 13.86 = ππ Be careful on this type of question. If particle is held by 2 different strings, the tensions may be different in each! The question asked for the weight, not the mass! (weight being mass x gravity…) 4B Statics of a Particle You need to know when to include additional forces on your diagrams, such as weight, tension, thrust, the normal reaction and friction A small bag of mass 10kg is attached at C to the ends of two light inextensible strings AC and BC. The other ends of the strings are attached to fixed points A and B on the same horizontal line. The bag hangs in equilibrium with AC and BC inclined to the horizontal at 30° and 60° respectively as shown. A B Draw a diagram T1 T1Sin30 T2 T2Sin60 30° C 60° T1Cos30 T2Cos60 10g Resolving Horizontally πΉ = ππ Calculate: a) The tension in AC b) The tension in BC π1 πΆππ 30 π2 = πΆππ 60 ο The strings are separate so use T1 and T2 as the tensions π2 πΆππ 60 − π1 πΆππ 30 = 0 π2 πΆππ 60 = π1 πΆππ 30 π1 πΆππ 30 π2 = πΆππ 60 Sub in values, choosing T2 as the positive direction Add T1Cos30 Divide by Cos60 4B Statics of a Particle You need to know when to include additional forces on your diagrams, such as weight, tension, thrust, the normal reaction and friction A small bag of mass 10kg is attached at C to the ends of two light inextensible strings AC and BC. The other ends of the strings are attached to fixed points A and B on the same horizontal line. The bag hangs in equilibrium with AC and BC inclined to the horizontal at 30° and 60° respectively as shown. Calculate: a) The tension in AC b) The tension in BC π1 πΆππ 30 π2 = πΆππ 60 A B Draw a diagram T1 T1Sin30 T2 ο The strings are separate so use T1 and T2 as the tensions T2Sin60 30° C 60° T1Cos30 T2Cos60 Resolving Vertically 10g πΉ = ππ π1 πππ30 + π2 πππ60 − 10π = 0 Sub in values, choosing T2 as the positive direction Replace T2 with the expression involving T1 π1 πΆππ 30 π1 πππ30 + πππ60 − 10π = 0 πΆππ 60 Multiply all terms by Cos60 π1 πππ30πΆππ 60 + π1 πΆππ 30πππ60 − 10ππΆππ 60 = 0 π1 = 49π Divide by the bracket π1 (πππ30πΆππ 60 + πΆππ 30πππ60) = 10ππΆππ 60 π1 = Calculate! Add 10gCos60 and factorise left side 10ππΆππ 60 πππ30πΆππ 60 + πΆππ 30πππ60 π1 = 49π 4B Statics of a Particle A You need to know when to include additional forces on your diagrams, such as weight, tension, thrust, the normal reaction and friction A small bag of mass 10kg is attached at C to the ends of two light inextensible strings AC and BC. The other ends of the strings are attached to fixed points A and B on the same horizontal line. The bag hangs in equilibrium with AC and BC inclined to the horizontal at 30° and 60° respectively as shown. Calculate: a) The tension in AC b) The tension in BC π2 = π1 πΆππ 30 πΆππ 60 π1 = 49π π2 = 84.87π B Draw a diagram T1 T1Sin30 T2 T2Sin60 30° C 60° T1Cos30 T2Cos60 ο The strings are separate so use T1 and T2 as the tensions 10g Find T2 by using the original equation… π2 = π2 = π1 πΆππ 30 πΆππ 60 49πΆππ 30 πΆππ 60 Sub in the value of T1 Calculate! π2 = 84.87π 4B Statics of a Particle R You need to know when to include additional forces on your diagrams, such as weight, tension, thrust, the normal reaction and friction A mass of 3kg rests on the surface of a smooth plane inclined at an angle of 45° to the horizontal. The mass is attached to a cable which passes up the plane and passes over a smooth pulley at the top. The cable carries a mass of 1kg which hangs freely at the other end. There is a force of PN acting horizontally on the 3kg mass and the system is in equilibrium. By modelling the cable as a light inextensible string and the masses as particles, calculate: a) The magnitude of P b) The normal reaction between the mass and the plane 9.8N T T 9.8N P PSin45 45Λ PCos45 3g 45Λ 3gCos45 45Λ 3gSin45 1g Find the tension using the 1kg mass πΉ = ππ π − 1π = 0 Resolve in the direction of T and sub in values Add 1g π = 1π π = 9.8π 4B Statics of a Particle R You need to know when to include additional forces on your diagrams, such as weight, tension, thrust, the normal reaction and friction A mass of 3kg rests on the surface of a smooth plane inclined at an angle of 45° to the horizontal. The mass is attached to a cable which passes up the plane and passes over a smooth pulley at the top. The cable carries a mass of 1kg which hangs freely at the other end. There is a force of PN acting horizontally on the 3kg mass and the system is in equilibrium. By modelling the cable as a light inextensible string and the masses as particles, calculate: a) The magnitude of P b) The normal reaction between the mass and the plane 9.8N 9.8N P PSin45 45Λ PCos45 3g 3gCos45 45Λ 45Λ 3gSin45 1g Resolve Parallel to find P πΉ = ππ ππΆππ 45 + 9.8 − 3ππππ45 = 0 Choose P as the positive direction and sub in values Rearrange ππΆππ 45 = 3ππππ45 − 9.8 3ππππ45 − 9.8 π= πΆππ 45 Divide by Cos45 Calculate π = 15.54π π = 15.54π 4B Statics of a Particle R You need to know when to include additional forces on your diagrams, such as weight, tension, thrust, the normal reaction and friction A mass of 3kg rests on the surface of a smooth plane inclined at an angle of 45° to the horizontal. The mass is attached to a cable which passes up the plane and passes over a smooth pulley at the top. The cable carries a mass of 1kg which hangs freely at the other end. There is a force of PN acting horizontally on the 3kg mass and the system is in equilibrium. By modelling the cable as a light inextensible string and the masses as particles, calculate: a) The magnitude of P b) The normal reaction between the mass and the plane π = 15.54π 9.8N 9.8N P PSin45 45Λ PCos45 3g 3gCos45 45Λ 45Λ 3gSin45 1g Resolve Perpendicular to find R πΉ = ππ π − ππππ45 − 3ππΆππ 45 = 0 Choose R as the positive direction and sub in values π = ππππ45 + 3ππΆππ 45 Rearrange Calculate π = 31.78π π = 31.78π 4B Statics of a Particle You can also solve statics problems by using the relationship F = µR We have seen before that FMAX is the maximum frictional force possible between two surfaces, and that it will resist any force up to this amount A block of mass 3kg rests on a rough horizontal plane. The coefficient of friction between the block and the plane is 0.4. When a horizontal force PN is applied to the block, the block remains in equilibrium. a) Find the value for P for which the equilibrium is limiting b) Find the value of F when P = 8N 3g R Remember that the frictional force can be lower than this and still prevent movement In statics, FMAX is reached when a body is in limiting equilibrium, ie) on the point of moving F Resolve vertically for R πΉ = ππ P 3kg Sub in values with R as positive 3g Find FMAX πΉππ΄π = ππ Sub in values π − 3π = 0 πΉππ΄π = (0.4)(3π) It is important to consider which Add 3g Calculate direction the object is about to move π = 3π πΉππ΄π = 11.76π as this affects the direction the friction is acting… So if P = 11.76N, then the block is in limiting equilibrium on the point of moving For part b), if P = 8N then equilibrium is not limiting, and P will be matched by a frictional force of 8N 4C Statics of a Particle You can also solve statics problems by using the relationship F = µR A mass of 8kg rests on a rough horizontal plane. The mass may be modelled as a particle, and the coefficient of friction between the mass and the plane is 0.5. Find the magnitude of the maximum force PN, which acts on this mass without causing it to move if P acts at an angle of 60° above the horizontal. πΉππ΄π = 4π − 0.5ππππ60 R F 8kg Draw a diagram P 60° PCos60 PSin60 ο Find the normal reaction as we need this for FMAX 8g Resolve Vertically πΉ = ππ π + ππππ60 − 8π = 0 Sub in values with R as positive π = 8π − ππππ60 Rearrange to find R in terms of P Find FMAX πΉππ΄π = ππ πΉππ΄π = (0.5)(8π − ππππ60) Sub in values Multiply bracket out πΉππ΄π = 4π − 0.5ππππ60 4C Statics of a Particle You can also solve statics problems by using the relationship F = µR A mass of 8kg rests on a rough horizontal plane. The mass may be modelled as a particle, and the coefficient of friction between the mass and the plane is 0.5. Find the magnitude of the maximum force PN, which acts on this mass without causing it to move if P acts at an angle of 60° above the horizontal. πΉππ΄π = 4π − 0.5ππππ60 R F 8kg Draw a diagram P 60° PCos60 PSin60 ο The horizontal forces will cancel out as the block is in limiting equilibrium 8g Resolve Horizontally πΉ = ππ ππΆππ 60 − πΉ = 0 ππΆππ 60 − (4π − 0.5ππππ60) = 0 ππΆππ 60 − 4π + 0.5ππππ60 = 0 Sub in values with P as positive Sub in FMAX ‘Multiply out’ the bracket Add 4g ππΆππ 60 + 0.5ππππ60 = 4π If P is any greater, the block will start to accelerate. If P is any smaller, then FMAX will be less and hence the block will not be in limiting equilibrium ο Find the normal reaction as we need this for FMAX π(πΆππ 60 + 0.5πππ60) = 4π Factorise P on the left side 4π π= πΆππ 60 + 0.5πππ60 π = 42.01π Divide by the bracket Calculate 4C Statics of a Particle R F You can also solve statics problems by using the relationship F = µR A box of mass 10kg rests in limiting equilibrium on a rough plane inclined at 20° above the horizontal. Find the coefficient of friction between the box and the plane. ο Draw a diagram 20° 10g 20° πΉππ΄π = π(10ππΆππ 20) 10gSin20 Resolving Perpendicular πΉ = ππ π − 10ππΆππ 20 = 0 ο We need to find FMAX so begin by calculating the normal reaction 10gCos20 Sub in values with R as positive Rearrange π = 10ππΆππ 20 Finding FMAX πΉππ΄π = ππ πΉππ΄π = π(10ππΆππ 20) Sub in R and leave µ 4C Statics of a Particle R F You can also solve statics problems by using the relationship F = µR A box of mass 10kg rests in limiting equilibrium on a rough plane inclined at 20° above the horizontal. Find the coefficient of friction between the box and the plane. ο Draw a diagram 20° 10g 20° 10gSin20 Resolving Parallel πΉ = ππ 10ππππ20 − πΉ = 0 ο We need to find FMAX so begin by calculating the normal reaction πΉππ΄π = π(10ππΆππ 20) ο Now you can resolve Parallel to find µ 10gCos20 10ππππ20 − π(10ππΆππ 20) = 0 Sub in values with ‘down the plane’ as positive Sub in FMAX 10ππππ20 = π(10ππΆππ 20) 10ππππ20 =π 10ππΆππ 20 Add µ(10gCos20) Divide by the bracket Calculate 0.36 = π 4C Statics of a Particle You can also solve statics problems by using the relationship F = µR A parcel of mass 2kg is placed on a rough plane inclined at an angle θ to the horizontal where Sinθ = 5/13. The coefficient of friction is 1/3. Find the magnitude of force PN, acting up the plane, that causes the parcel to be in limiting equilibrium and on the point of: a) 5 13 Moving up the plane πΆππ π = 12 13 ππππ = Hyp 5 12 13 5 Opp θ ππππ = πππ π»π¦π 5 ππππ = 13 b) Moving down the plane ππππ = Find the other trig ratios – this will be useful later! 12 Adj So the opposite side is 5 and the hypotenuse is 13 ο Use Pythagoras to find the missing side! ο Now you can work out the other 2 trig ratio… πΆππ π = π΄ππ π»π¦π ππππ = πππ π΄ππ πΆππ π = 12 13 ππππ = 5 12 4C Statics of a Particle R P You can also solve statics problems by using the relationship F = µR A parcel of mass 2kg is placed on a rough plane inclined at an angle θ to the horizontal where Sinθ = 5/13. The coefficient of friction is 1/3. Find the magnitude of force PN, acting up the plane, that causes the parcel to be in limiting equilibrium and on the point of: a) πΉππ΄π = 5 13 πΆππ π = 2 ππΆππ π 3 12 13 ππππ = F θ 2gCosθ 2gSinθ Resolving Perpendicular for R πΉ = ππ π − 2ππΆππ π = 0 Moving up the plane Sub in values with R as the positive direction π = 2ππΆππ θ b) Moving down the plane ππππ = 2g θ 5 12 Start with a diagram ο P is acting up the plane, on the point of causing the box to move ο Friction is opposing this movement Rearrange Finding FMAX πΉππ΄π = ππ πΉππ΄π 1 = (2ππΆππ π) 3 πΉππ΄π 2 = ππΆππ π 3 Sub in values Remove the bracket 4C Statics of a Particle R P You can also solve statics problems by using the relationship F = µR A parcel of mass 2kg is placed on a rough plane inclined at an angle θ to the horizontal where Sinθ = 5/13. The coefficient of friction is 1/3. Find the magnitude of force PN, acting up the plane, that causes the parcel to be in limiting equilibrium and on the point of: a) πΉππ΄π = 5 13 πΆππ π = 2 ππΆππ π 3 12 13 ππππ = F θ 2gSinθ Resolving Parallel for P πΉ = ππ Sub in values with P as the positive direction π − 2πππππ − πΉ = 0 Moving up the plane π = 13.57π Sub in F 2 π − 2πππππ − ππΆππ π = 0 3 b) Moving down the plane ππππ = 2gCosθ 2g θ 5 12 Start with a diagram ο P is acting up the plane, on the point of causing the box to move ο Friction is opposing this movement 2 π = 2πππππ + ππΆππ π 3 π = 2π 5 2 12 + π 13 3 13 π = 13.57π Rearrange for P Sub in Sinθ and Cosθ Calculate 4C Statics of a Particle F P R You can also solve statics problems by using the relationship F = µR A parcel of mass 2kg is placed on a rough plane inclined at an angle θ to the horizontal where Sinθ = 5/13. The coefficient of friction is 1/3. Find the magnitude of force PN, acting up the plane, that causes the parcel to be in limiting equilibrium and on the point of: a) Moving up the plane π = 13.57π b) Moving down the plane π = 1.51π 12 5 5 πΆππ π = ππππ = ππππ = 13 13 12 πΉππ΄π = 2 ππΆππ π 3 2gCosθ 2g θ F θ 2gSinθ Resolving Parallel for P πΉ = ππ We now need to adjust the diagram for part b) ο Now, as the particle is on the point of sliding down the plane, the friction will act up the plane instead… ο FMAX will be the same as before as we haven’t changed any vertical components Sub in values with P as the positive direction π + πΉ − 2πππππ = 0 Replace F 2 π + ππΆππ π − 2πππππ = 0 3 2 π = 2πππππ − ππΆππ π 3 π = 2π 5 2 12 − π 13 3 13 π = 1.51π Rearrange Sub in Sinθ and Cosθ Calculate 4C Statics of a Particle F P R You can also solve statics problems by using the relationship F = µR A parcel of mass 2kg is placed on a rough plane inclined at an angle θ to the horizontal where Sinθ = 5/13. The coefficient of friction is 1/3. Find the magnitude of force PN, acting up the plane, that causes the parcel to be in limiting equilibrium and on the point of: a) Moving up the plane π = 13.57π b) Moving down the plane π = 1.51π 2g θ θ 2gCosθ 2gSinθ We now need to adjust the diagram for part b) ο Now, as the particle is on the point of sliding down the plane, the friction will act up the plane instead… ο FMAX will be the same as before as we haven’t changed any vertical components A force of 13.57N up the plane is enough to bring the parcel to the point of moving in that direction. Any more will overcome the combination of gravity and friction and the parcel will start moving up A force of 1.51N up the plane is enough, when combined with friction, to prevent the parcel from slipping down the plane and hold it in place. Any less and the parcel will start moving down. 4C Statics of a Particle 15N You can also solve statics problems by using the relationship F = µR A box of mass 1.6kg is placed on a rough plane, inclined at 45° to the horizontal. The box is held in equilibrium by a light inextensible string, which makes an angle of 15° with the plane. When the tension in the string is 15N, the box is in limiting equilibrium and about to move up the plane. Find the value of the coefficient of friction between the box and the plane. πΉππ΄π = π(1.6ππΆππ 45 − 15πππ15) Draw a diagram – ensure you include all forces and their components in the correct directions R 15° 1.6g F 45° 45° 1.6gCos45 ο The box is on the point of moving up, so friction is acting down the plane 1.6gSin45 ο Find the normal reaction and use it to find FMAX Resolving Perpendicular πΉ = ππ π + 15πππ15 − 1.6ππΆππ 45 = 0 Sub in values with R as the positive direction Rearrange π = 1.6ππΆππ 45 − 15πππ15 Finding FMAX πΉππ΄π = ππ Sub in values πΉππ΄π = π(1.6ππΆππ 45 − 15πππ15) 4C Statics of a Particle 15N You can also solve statics problems by using the relationship F = µR A box of mass 1.6kg is placed on a rough plane, inclined at 45° to the horizontal. The box is held in equilibrium by a light inextensible string, which makes an angle of 15° with the plane. When the tension in the string is 15N, the box is in limiting equilibrium and about to move up the plane. Draw a diagram – ensure you include all forces and their components in the correct directions R 15° 1.6g F 45° 45° ο Now resolve parallel to create an equation you can solve for μ. 1.6gCos45 1.6gSin45 Resolving Parallel πΉ = ππ Find the value of the coefficient of 15πΆππ 15 − 1.6ππππ45 − πΉ = 0 friction between the box and the 15πΆππ 15 − 1.6ππππ45 − π(1.6ππΆππ 45 − 15πππ15) = 0 plane. πΉππ΄π = π(1.6ππΆππ 45 − 15πππ15) Divide by the bracket Calculate! Sub in values with ‘up’ the plane as the positive direction Replace F 15πΆππ 15 − 1.6ππππ45 = π(1.6ππΆππ 45 − 15πππ15) Add μ term 15πΆππ 15 − 1.6ππππ45 =π 1.6ππΆππ 45 − 15πππ15 0.47 = π 4C Statics of a Particle You can also solve statics problems by using the relationship F = µR A box of mass 1.6kg is placed on a rough plane, inclined at 45° to the horizontal. The box is held in equilibrium by a light inextensible string, which makes an angle of 15° with the plane. When the tension in the string is 15N, the box is in limiting equilibrium and about to move up the plane. Find the value of the coefficient of friction between the box and the plane. ο The tension is reduced to 10N. Determine the magnitude and direction of the frictional force in this case 10N 15N Update the diagram (or re-draw it!) 15° ο Calculate the new FMAX, first finding the new R… R π = 0.47 1.6g F 45° 45° 1.6gCos45 1.6gSin45 Resolving Perpendicular πΉ = ππ π + 10πππ15 − 1.6ππΆππ 45 = 0 Sub in values with R as the positive direction Rearrange π = 1.6ππΆππ 45 − 10πππ15 Finding FMAX πΉππ΄π = ππ Sub in values πΉππ΄π = 0.47(1.6ππΆππ 45 − 10πππ15) Calculate πΉππ΄π = 4.012π 4C Statics of a Particle You can also solve statics problems by using the relationship F = µR A box of mass 1.6kg is placed on a rough plane, inclined at 45° to the horizontal. The box is held in equilibrium by a light inextensible string, which makes an angle of 15° with the plane. When the tension in the string is 15N, the box is in limiting equilibrium and about to move up the plane. Find the value of the coefficient of friction between the box and the plane. ο The tension is reduced to 10N. Determine the magnitude and direction of the frictional force in this case 10N Update the diagram (or re-draw it!) 15° ο Calculate the new FMAX, first finding the new R… R π = 0.47 1.6g F 45° 45° 1.6gCos45 1.6gSin45 πΉππ΄π = 4.012π ο Add up the forces acting parallel to the plane (ignoring friction for now) Resolving Parallel (without friction) The force up the plane will be given by: 10πΆππ 15 − 1.6ππππ45 = −1.428π As this is negative, then without friction, there is an overall force of 1.428N acting down the plane ο Therefore, friction will oppose this by acting up the plane ο As FMAX = 4.012N, the box will not move and is not in limiting equilibrium 4C Summary • We have learnt about resolving forces when a particle is in limiting equilibrium • We have seen when and how to include additional forces such as tension and friction • We have looked at situations where friction acts in different directions
