©CourseCram PHYS117 MIDTERM SOLUTIONS 2019 PRACTICE 1 2019 PRACTICE 1 (note that this official βpracticeβ exam is mainly assembled from real past midterm and final exam questions, so itβs great practice!) Q1 C) DISPLACEMENT We can see the components of the final displacement ππ₯ = (β3) + 6 = 3 ππππππ ππ¦ = 4 ππππππ so the final displacement is |π| = βππ₯2 + ππ¦2 = β32 + 42 = 5 ππππππ π = tanβ1 ( ππ¦ 4 ) = tanβ1 ( ) = 53° ππ₯ 3 From the diagram we see this is north of east. The answer is C) 2 A) AVERAGE SPEED, PROJECTILE Note that we are asked for the average speed. The average velocity of the motion would be zero, as it ends up back where it started. For speed it starts at 29.4 m/s At the highest point it has a vertical speed of 0 m/s, and this is the only direction its moving so the speed is 0 m/s it comes back down to the ground at the same speed as it launched, by symmetry, so the speed is 29.4 m/s So the average speed when itβs going up is |π£ππ£π πππππ π’π | = π ππππ ππ‘ π π‘πππ‘ + π ππππ ππ‘ π‘ππ |π£0 | + 0 29.4 = = = 14.7 π/π 2 2 2 Likewise when coming down |π£ππ£π ππππππ πππ€π | = (π ππππ ππ‘ π‘ππ) + (π ππππ ππ‘ πππ) 0 + |π£0 | 29.4 = = = 14.7 π/π 2 2 2 It spends the same amount of time going up and coming down, so |π£ππ£π π‘ππ‘ππ | = 14.7 + 14.7 = 14.7 π/π 2 1 ©CourseCram Q3 D) PHYS117 MIDTERM SOLUTIONS 2019 PRACTICE 1 FORCES, FRICTION, NORMAL FORCE This is an βall for aβ problem: when we have a set of blocks all moving together, we can find the acceleration π of all of them by looking at all of them, then find the value we want. Looking at the whole assembly, the total frictional force will be 3 × 3 = 9 π The total mass is 3 × 3 = 9 ππ So in the x-direction Ξ£πΉπ₯ = ππ‘ π +18 β 9 = 9π π = 1 π/π 2 We want to find the normal force exerted on the middle cube by the left cube. So we must draw a diagram not including the left cube, so that its force is external on our diagram. Itβs simplest to look at the middle and right cube together. That way the force between the middle and right cubes is internal and doesnβt have to be included. This leaves us with a mass of two cubes π2 = 2 × 3 = 6 ππ The total frictional force due to 2 cubes is 2 × 3 = 6 π to the left We now have Ξ£πΉπ₯ = π2 π +πππππ‘ β 6 = 6(1) πππππ‘ = 12 π 2 ©CourseCram Q4 A) PHYS117 MIDTERM SOLUTIONS 2019 PRACTICE 1 NORMAL FORCE, ELEVATOR The reading on the scale is the normal force. The person is riding up, and the elevator is slowing down, so the acceleration is opposite to the direction of motion β it is downwards, and ππ¦ = β2.5 π/π 2 Ξ£πΉπ¦ = ππ +π β ππ = ππ π β 70(9.8) = 70(β2.5) π = 511 β 510 π Q5 D) RELATIVE MOTION The boat tries to move 8.0 m/s north, but the current washes it east at 2.0 m/s. So the resultant velocity is |π£πππ π’ππ‘ | = β82 + 22 = 8.2 π/π 2 π = tanβ1 ( ) = 14° πΈππ π‘ ππ ππππ‘β 8 So this angle is 90 β 14 = 76° ππππ‘β ππ πππ π‘ 3 ©CourseCram Q6 A) PHYS117 MIDTERM SOLUTIONS 2019 PRACTICE 1 PROJECTILE MOTION The speed is the total speed due to both x and y components |π£πππππ | = βπ£π₯2 + π£π¦2 The wheel starts with the same velocity as the helicopter π£0π₯ = 54 π/π π£0π¦ = 0 π/π There is no horizontal acceleration so the horizontal component does not change π£π₯ = π£0π₯ = 54 π/π The vertical component when hitting the ground can be found by the equations of motion 2 π£π¦2 = π£0π¦ + 2π(π¦ β π¦0 ) π£π¦2 = 0 + 2(β9.8)(0 β 100) π£π¦2 = 1960 Taking the square root can give a positive or negative answer. We know that this component will be negative. π£π¦ = β44.3 π/π and so |π£| = β542 + (β44.3)2 = 69.8 π/π β 70 π/π 4 ©CourseCram Q7 A) PHYS117 MIDTERM SOLUTIONS 2019 PRACTICE 1 FRICTION (on a slope) The block does NOT slide, so the force definitely doesnβt involve kinetic friction, so the answer is NOT B). The force holding the block in place is the force of static friction. But this is NOT necessarily the maximum force of static friction! Weβre not told that the block is βabout to slideβ or that weβre βat the steepest angleβ, so we can NOT say anything about ππ . Remember that 0 < ππ < ππ πππ₯ = ππ π Looking at the free body diagram The block is not sliding along the tilted x-axis, so Ξ£πΉπ₯ = 0 βππ + ππ sin π = 0 ππ = ππ sin π So the answer is A) BONUS: If weβd been told βthe slope is increased until the block is just about to start slidingβ, THEN we could say the static friction was maximum, and ππ = ππ πππ₯ = ππ π = ππ ππ cos π 5 ©CourseCram Q8 D) PHYS117 MIDTERM SOLUTIONS 2019 PRACTICE 1 FRICTION The sled is being pulled at a constant speed in a constant direction, so it has constant velocity, and so the acceleration is zero. So the total force is zero. Weβre doing friction so we look in the N direction first Ξ£πΉπ¦ = 0 +π β π cos π = 0 π β 150 cos 28 = 0 π = 132.4 πππ€π‘πππ then Ξ£πΉπ₯ = 0 β120 + π sin π + ππ = 0 β120 + 150 sin 28 + ππ π = 0 β120 + 150 sin 28 + ππ (132.4) = 0 ππ = 0.37 6 ©CourseCram Q9 D) PHYS117 MIDTERM SOLUTIONS 2019 PRACTICE 1 SPRING FORCE, SLOPE The reading in the spring scale is the force in the spring. The system is in equilibrium so Ξ£πΉ = 0 Looking at the block Ξ£πΉπ₯ = 0 βππ sin π + πΉπ = 0 β10(9.8) sin 60 + πΉπ = 0 πΉπ = 84.9 π Q10 C) TENSION, FRICTION. We have two blocks and we want to find forces involved in them. Itβs easiest to look at each block. The 5 kg block is moving to the right with an acceleration π5 = +π the hanging 9 kg weight is falling with the same size of acceleration downwards, π9 = βπ The hanging weight is simpler so we look at that first. Ξ£πΉπ¦ = π9 π9 +π β π9 π = π9 (βπ) π β 9(9.8) = 9(βπ) One equation with two unknowns, so we rewrite as one unknown equals everything else π = 88.2 β 9π Looking at the 5 kg block 7 ©CourseCram PHYS117 MIDTERM SOLUTIONS 2019 PRACTICE 1 We have friction so we look at the normal force direction first Ξ£πΉπ¦ = 0 +π β π5 π = 0 π β 5(9.8) = 0 π = 49 πππ€π‘πππ then Ξ£πΉπ₯ = π5 π5 +π β ππ = π5 (+π) But we know the tension (88.2 β 9π) β ππ π = π5 π 88.2 β 9π β 0.1(49) = 5π 83.3 = 14π π = 5.95 π/π 2 and so π = 88.2 β 9(5.95) = 34.65 β 35 π 8 ©CourseCram Q11 C) PHYS117 MIDTERM SOLUTIONS 2019 PRACTICE 1 NEWTONβS THIRD LAW By Newtonβs third law if the daughter exerts a force on the father, the father exerts exactly the same force magnitude on the father. So the answer cannot be A) or B) The effect of the force is given by newtonβs second law πΉ = ππ If the father weights twice as much as the daughter, he has twice the mass. And from π= πΉ π if the daughter has half the mass, she will experience twice the acceleration. The answer is C Q12 A) FORCES, WEIGHT, FRICTION The force of kinetic friction is constant. The slope is constant. Everything is constant, and so the acceleration is constant, and sot he object will continue to speed up at a constant rate. 9 PREP101 PHYS117 MIDTERM SOLUTIONS 2019 PRACTICE 2 MCQ 2019 PRACTICE 2 MCQ (note that this βpractice examβ is made entirely of REAL past PHYS117 final exam questions, from 2011 and 2007, so it is certainly excellent practice!) Q1 C) CIRCULAR MOTION For orbital motion, the net force towards the center of the circle is the centripetal force. Here Ξ£πΉπ = πΉπΆ πΊππ ππ£ 2 = π2 π πΊπ = π£2 π πΊπ π£=β π = β( 6.67 × 10β11 × 5.68 × 1026 ) 100,000 × 103 = 1.95 x 104 m/s = 19.5 km/s Q2 D) WORK, KINETIC ENERGY By conservation of energy πΈπ + πππ = πΈπ πΎπΈπ + ππΈππ + πππ = πΎπΈπ + ππΈππ So if we want the change in kinetic energy we rearrange to get πΎπΈπ β πΎπΈπ = πππ + ππΈππ β ππΈππ We set the original height to zero, so πΎπΈπ β πΎπΈπ = πππ β ππΈππ The work done by non-conservative forces is the work done by the 150 N force and the work done by friction πππ = π150 + ππ π150 = πΉπ = 150(12) = 1800 π½ ππ = βππ 10 PREP101 PHYS117 MIDTERM SOLUTIONS 2019 PRACTICE 2 MCQ = βπππ = βπππ cos π π = β0.1(20)(9.8) cos 30 (12) = β203.7 π½ So we have πππ = 1800 + (β203.7) = 1596.3 and so πΎπΈπ β πΎπΈπ = 1596.3 β ππβ = 1596.3 β 20(9.8)(12 sin 30) = 420 π½ ALTERNATE METHOD Work energy theorem Work done = change in energy Wnc = Ef - Ei W150 - Wfriction = ΞKE + ΞPE ΞKE = 150(12) β Ffr(12) - ΞPE We have Wfriction = Ffr d = ΞΌN d = ΞΌmgcosΞΈ d = 0.1(20)(9.80)cos30(12) = 203.7 J and ΞPE = mgh = 20 (9.80) 12sin30 = 1176 J ΞKE = 150(12) β 203.7 - 1176 = 420 J 11 PREP101 Q3 D) PHYS117 MIDTERM SOLUTIONS 2019 PRACTICE 2 MCQ CONSERVATION OF ENERGY, CIRCULAR MOTION The ball is going in a circle so the force towards the center must always equal Fc=mv2/r If the string is just short of going slack at the top, the tension is zero. and we have Ξ£Fy = -mg = -mv2/L vtop = β(Lg) Moving from the top to the bottom β weβre talking about changing velocities and heights, so itβs conservation of energy Etop = Ebottom KE + PE = KE ½ mvtop2 + mg2L = ½ mvbottom2 ½ (Lg) + 2Lg = ½ vb2 vb = β(5Lg) 12 PREP101 Q4 C) PHYS117 MIDTERM SOLUTIONS 2019 PRACTICE 2 MCQ Scaling (with Gravitational Force) A scaling question, so the fastest method is ratios! πππ‘ππ = πππ€ πππ πΊπ π π πΉππ ( ππ2 ) = πΊππΈ π πΉπ 2 ππΈ πΊ( πΉππ = 600 ( 1 )ππΈ π 100 2 1 ( ππΈ ) 4 ) πΊππΈ π ππΈ2 πΉππ = 600 1 100 1 2 (4) πΉππ = 96 π ALTERNATE METHOD Writing the old value then the new one On earth, his weight is π = πΉπ = πΊππΈ π = 600 ππΈ2 On the other planet, he weighs ππ = πΉππ = = πΊππ π ππ2 πΊ(0.01ππΈ )π (0.25ππΈ )2 = 0.16 πΊππΈ π ππΈ2 = 0.16(600) = 96 πππ€π‘πππ Q5 A) Impulse βAverage forceβ means itβs an impulse question I = FavgΞt = Ξp 5000Ξt = pnew - pold 5000Ξt = 0.14(37) β 0.14(-42.3) Ξt = 2.22 x 10-3 s 13 PREP101 Q6 A) PHYS117 MIDTERM SOLUTIONS 2019 PRACTICE 2 MCQ Momentum (2D) Explosion means conservation of momentum. Choose the 1kg piece to be shooting straight up. Then we have px before= px after 0 = -4v4cos30 + 5v5cos 30 v5 = 0.8 v4 py before= py after 0 = + 1(10) β 4v4sin30 β 5v5sin30 0 = 10 β 2 v4 β 2.5(0.8)v4 10 = 4v4 v4= 2.5 m/s v5 = 0.8(2.5) = 2 m/s 14 PREP101 Q7 E) PHYS117 MIDTERM SOLUTIONS 2019 PRACTICE 2 MCQ Power (Ski lift) Power = W/t W=E The energy given to each passenger is E = KE + PE = ½ mv2 + mgh = ½ 190 (6)2 + 190 (9.80) 150 = 2.83 x 105 J Then the power is P = 2.83 x 105/30 = 9.43 x 103 W A difference of only 0.01 is usually a rounding error. The answer is E) Q8 A) Equations of motion The car is moving at v0 = 25 m/s, and decelerates at a = - 8.0 m/s2 to final velocity of v = 0 m/s. We can use the equation of motion without time π£ 2 = π£02 + 2ππ₯π₯ 0 = 252 + 2(β8)π₯π₯ π₯π₯ = 39.1 π 15 PREP101 Q9 B) PHYS117 MIDTERM SOLUTIONS 2019 PRACTICE 2 MCQ Spring Potential Energy Comparing an old and a new value, so a quick way of doing this is with ratios πππ‘ππ = πππ€ πππ 1 2 ππΈπ πππ€ 2 ππ₯π =1 ππΈπ πππ ππ₯π2 2 100 π₯π2 = (0.12)2 72 π₯π = 0.141 π = 14.1 ππ ALTERNATE METHOD We can also do this by writing down the old value, then the new one The energy stored in a spring is given by 1 ππΈπ = ππ₯π₯ 2 2 If 72 J is stored by 0.12 m of compression 1 72 = π(0.12)2 2 72 = 0.0072 π π = 10,000 π/π So to store 100 J, we require 1 ππΈπ = ππ₯π₯ 2 2 1 100 = (10,000)π₯π₯ 2 2 100 = 5,000 π₯π₯ 2 π₯π₯ = 0.141 π 16 PREP101 Q10 C) PHYS117 MIDTERM SOLUTIONS 2019 PRACTICE 2 MCQ Power The average power is given by π= π π₯πΈ = π‘ π‘ Here the change in energy is the change in kinetic energy π₯πΈ = πΎπΈπ β πΎπΈπ 1 = ππ£ 2 β 0 2 1 = 1000(25)2 2 = 312,500 = 313 ππ½ And the power is π= = π₯πΈ π‘ 312,500 7.5 = 41,666 π = 41.7 ππ Q11 E) Impulse We have the length of time of the collision, and are asked for the "average force" - these are keywords which tell us it is an impulse question πΌ = π₯π = πΉππ£πππππ π₯π‘ ππ β ππ = πΉππ£π π₯π‘ ππ£π β ππ£π = πΉππ£π π₯π‘ Remember to include the signs of the directions for velocity - up is positive, down is negative 0.3(4.2) β 0.3(β4.5) = πΉππ£π 0.03 πΉππ£π = 87 π 17 PREP101 Q12 D) PHYS117 MIDTERM SOLUTIONS 2019 PRACTICE 2 MCQ Conservation of momentum, Forces, Equations of motion A block grinding to a halt over friction is a classic conservation of energy problem πΈπ + ππβπ = πΈπ πΎπΈπ + ππΈπ + ππβπ = πΎπ + ππΈπ There are no heights or springs so the potential energies are both zero. It grinds to a halt so the final kinetic energy is zero. πΎπΈπ + ππβπ = 0 The non-conservative force is zero, so we have ππβπ = βππ π = βππ ππ = βππ πππ and we have 1 ππ£π2 β ππ πππ = 0 2 1 2 π£ β ππ ππ = 0 2 π 1 2 π£ β 0.5(9.8)(2.1) = 0 2 π π£π = 4.54 π/π This is the initial speed for the energy section of the problem, so itβs the final speed for the next part of the problem, the collision. Then we can look at the collision. The large lump of clay starts from rest, and so has zero velocity and momentum πππππ‘πππ = ππππππ 0.1π£1π + 0 = (0.1 + 0.5)4.54 π£1π = 27.2 π/π ALTERNATE METHOD We can also work out the speed from forces and equations of motion, but as we see it takes longer When the two lumps of clay stick together, that is a perfectly inelastic collision, so that is conservation of momentum. 18 PREP101 PHYS117 MIDTERM SOLUTIONS 2019 PRACTICE 2 MCQ The resulting lump is moving and then comes to a halt, so that is equations of motion. The force that halts it is friction, so that is forces. π΄πΉ = ππ βππ = (0.1 + 0.5)π βππ π = 0.6π The clay is on a horizontal surface with no angled forces, so N = mg βππ ππ = 0.6π β0.5(0.6)9.8 = 0.6π π = β4.9 π/π 2 Now we can look at the lump grinding to a halt. We are given distances and no time, so we use the equation of motion without time. π£ 2 β π£02 = 2ππ₯π₯ 02 β π£02 = 2(β4.9)(2.1) π£0 = 4.54 π/π 19 PREP101 Q13 D) PHYS117 MIDTERM SOLUTIONS 2019 PRACTICE 2 MCQ Centripetal force Because the object is moving in a circle, the total force towards the center of the circle must equal the centripetal force. At the bottom of the circle, the center is above it, in the +y direction, so we have π΄πΉπ¦ = +πΉπ ππ£ 2 π β ππ = π π = ππ + = 0.4(9.8) + ππ£ 2 π 0.4(4)2 0.5 π = 16.7 π 20 PREP101 Q14 A) PHYS117 MIDTERM SOLUTIONS 2019 PRACTICE 2 MCQ Scaling (with Gravitational Force) We are asked to find the new acceleration due to gravity, given the new planet's values compared to Earth. This is a classic ratio question. The gravitational acceleration is given by π= πΊπ = 9.8 π2 where M and r are the mass and radius of Earth. First we write down the values we're given: The new planet has three times the radius of Earth, so we have ππππ€ = 3π The new planet has the same average density as Earth, so we have ππππ€ = ππΈπππ‘β ππππ€ 4 3 πππππ€ 3 =4 3 π ππ 3 ππππ€ π = (3π)3 π 3 ππππ€ =π 27 ππππ€ = 27π Now we can work out the new acceleration due to gravity using the ratio ππππ€ = π πΊππππ€ 2 ππππ€ πΊπ π2 ππππ€ = 9.8 ππππ€ 2 ππππ€ π π2 ππππ€ = 9.8 27π (3π)2 π π ππππ€ 27 = 9.8 9 ππππ€ = 29.4 π/π 2 21 PREP101 PHYS117 MIDTERM SOLUTIONS 2019 PRACTICE 2 long answer 2019 PRACTICE 2 long answer 1 Spring Forces, Conservation of Energy, Momentum a) A force of 2 N is required to compress the spring by 0.25 cm = 0.0025 m So we have |πΉπ | = |ππ₯π₯| 2 = π(0.0025) π = 800 π/π b) The bullet and block are moving at one point, and later have compressed the spring and stopped moving. This is two different locations with no time mentioned - this is a conservation of energy question. Kinetic energy has been converted into spring potential energy. There are no frictional or other external forces, so there is no external work done. πΈπ + ππβπ = πΈπ πΈπ = πΈπ πΎπΈπ = ππΈπ π 1 1 ππ£ 2 = π(π₯π₯)2 2 2 ππ£ 2 = π(π₯π₯)2 (0.01 + 0.99)π£ 2 = 800(0.15)2 π£ 2 = 18 π£ = 4.24 π/π c) We have objects colliding and sticking together, so we are dealing with perfectly inelastic collisions. πππππ‘πππ = ππππππ The block is initially not moving, so its initial momentum is zero. πππ’ππππ‘ π£0 + 0 = (πππ’ππππ‘ + ππππππ )π£ 0.01π£0 = (0.01 + 0.99)4.24 π£0 = 424 π/π 22 PREP101 PHYS117 MIDTERM SOLUTIONS 2019 PRACTICE 2 long answer 2 CONSERVATION OF ENERGY, CONSERVATION OF MOMENTUM a) C of E The block changes speed over a certain distance β perfect for conservation of energy or equation of motion. Conservation of energy is always faster πΈππππ‘πππ + ππβπ = πΈπππππ The initial energy is all kinetic, the work done by friction is negative, and the final energy is zero 1 ππ£ 2 + (βππ π) = 0 2 The force of friction is ππ = ππ π = ππ ππ And we have 1 ππ£ 2 β ππ πππ = 0 2 1 2 π£ β ππ ππ = 0 2 1 2 π£ β 0.25(9.8)(9.5) = 0 2 v = 6.82 m/s b) C of MOMENTUM Things colliding β this is a momentum question pbefore= pafter mbulletvbullet +mblock(0) = (mbullet + mblock)6.82 0.015vbullet = 1.115(6.82) vbullet = 507 m/s 23 PREP101 Q3 PHYS117 MIDTERM SOLUTIONS 2019 PRACTICE 2 long answer CIRCULAR MOTION Circular motion β so the net force towards the center of the circle is mv2/r. Here βtowards the center of the circleβ is in the negative x direction. We are told the length of the lower rope is 1 m, so we can find the radius of the circle from r = 1 cos30 Draw an FBD. In a rope question like this itβs fastest to do the y direction first Ξ£Fy= 0 Tsin60 + Tsin30 β mg = 0 1.366π β 2.5(9.8) = 0 π = 17.94 π Moving on to x Ξ£πΉπ₯ = ππ£ 2 π 2.5π£ 2 π cos 60 + π cos 30 = 1 cos 30 1.366(17.94) = 2.5π£ 2 cos 30 π£ = 2.91 π/π 24 PHY117 Midterm Solutions Q1 B) 2011 Q1 Equations of motion The distance is how much further the cheetah will travel than the gazelle in 30 seconds. We find distance from π·ππ π‘ππππ = π ππππ × π‘πππ but we need the speed in the same units as the time. We convert from km/hr to m/s by ππ/βπ = 1000 π 1000 1 = π/π = π/π 60(60) π 3600 3.6 The cheetah travels ππβπππ‘πβ = 100 × 30 = 833.3 π 3.6 ππππ§ππππ = 80 × 30 = 666.7 π 3.6 The gazelle travels The difference is 833.3 β 666.7 = 166.6 π β 167 π ALTERNATE METHOD The cheetah is moving 100 - 80 = 20 km/hr faster than the gazelle. So the difference in distance over 30 seconds is π= 20 × 30 = 166.7 π β 167 π 3.6 25 PHY117 Midterm Solutions Q2 E) 2011 Q2 Projectile There are two ways to do this, a fast way and a long way. The fast method is to remember that projectile motion is symmetric: the rock will come down with the same speed as at goes up. So when we throw a rock up at +15 m/s, it will fall back past us at -15 m/s. We only need to work out when that happens, and if we throw the second rock at the same time with the same velocity they will hit the ground at the same time. When will the upwards-thrown rock pass the top of the building? We find that from 1 Ξπ¦ = π£0π¦ π‘ β ππ‘ 2 2 1 π¦ β π¦0 = π£0π¦ π‘ β ππ‘ 2 2 1 100 β 100 = 15π‘ β (9.8)π‘ 2 2 0 = 15π‘ β 4.9π‘ 2 We don't need to do quadratic - we can divide by t to find 0 = 15 β 4.9π‘ 4.9π‘ = 15 π‘= 15 = 3.06 π 4.9 ALTERNATE METHOD The simpler, but longer method is to find the time taken for both rocks to hit the ground then find the difference. For the upward rock, the rock starts at y0 = 100 and an initial vertical velocity of v0y = 15 m/s. It hits the ground when y = 0. 1 Ξπ¦ = π£0π¦ π‘ β ππ‘ 2 2 Remember Ξπ¦ = π¦ β π¦0 which in this case is Ξπ¦ = 0 β 100 1 2 0 β 100 = (+15)π‘π’π β (9.8)π‘π’π 2 2 4.9π‘π’π β 15π‘π’π β 100 = 0 Solving for tup and taking the positive value, we find π‘π’π = 6.30 π Then we do the same thing for the second rock, which starts with (-15 m/s) 26 PHY117 Midterm Solutions 2011 Q2 1 Ξπ¦ = π£0π¦ π‘ β ππ‘ 2 2 1 2 0 β 100 = (β15)π‘πππ€π β (9.8)π‘πππ€π 2 2 4.9π‘πππ€π + 15π‘πππ€π β 100 = 0 π‘πππ€π = 3.24 π So the difference, how much later you should throw the second rock, is π‘π’π β π‘πππ€π = 6.30 β 3.24 = 3.06 π As you can see, this method is much longer, but it gives the same answer. 27 PHY117 Midterm Solutions Q3 C) 2011 Q3 Equations of motion This motion has two stages: in the first stage it is accelerating upwards, and in the second it continues in free-fall under the influence of gravity until it reaches its maximum height. Stage 1: Accelerating The rocket is accelerating at 30 m/ss for 1.0 s. We are told it is launched from the ground, so assume its initial velocity was zero. The height reached is 1 Ξπ¦ = π£0π¦ π‘ + ππ‘ 2 2 1 Ξπ¦1 = 0 + 30(1)2 2 Ξπ¦1 = 15 π At this point it has a new velocity, given by π£1 = π£0 + ππ‘ = 0 + 30(1) = 30 π/π Stage 2: Free-fall Even though the rocket is in "free-fall", it has a large upward velocity of 30 m/s, and will continue moving up until acceleration due to gravity causes it to start falling down. The maximum height of any projectile is given by the point where the vertical velocity is zero, so we use the formula π£ 2 = π£02 + 2πΞπ₯ The starting velocity of stage 2 is v1, the final velocity is zero, the change in position is the change in height over the second stage, Ξπ¦2 , and the acceleration is (-g), so we have π£ 2 = π£12 + 2(βπ)Ξπ¦2 0 = 302 + 2(β9.8)Ξπ¦2 Ξπ¦2 = 45.9 π So the total height covered is π¦ = Ξπ¦1 + Ξπ¦2 = 15 + 45.9 = 60.9 β 61 π 28 PHY117 Midterm Solutions Q4 A) 2011 Q4 Pythagoras Moving east-west and north-south, like moving horizontally and vertically, are two components of the total displacement. So we can use Pythagoras to find the total distance. The taxi moves a total of 5 blocks east, and a total of 7 blocks north, so we have πππ π‘ππππ = β52 + 72 = 8.6 ππππππ 29 PHY117 Midterm Solutions Q5 D) 2011 Q5 Projectile Motion Projectile motion is just the equations of motion in both x and y. To find out about the impact of a projectile, we could use the y-component equations of motion to find the time of impact, then the time to find out whatever else we want. But there's a faster way! We only want to know the velocity at impact, and we already have an equation to find the velocity at another location. 2 π£π¦2 = π£0π¦ + 2πΞπ¦ The launch vertical velocity component is π£0π¦ = π£0 sin π = 15 sin 53 The acceleration is downwards due to gravity π = β9.8 And the distance is Ξπ¦ = π¦ β π¦0 = 0 β 35 = β35 So we have π£π¦2 = (15 sin 53)2 + 2(β9.8)(β35) = 829.5 π£π¦ = 28.8 π/π β 29 π/π 30 PHY117 Midterm Solutions Q6 C) 2011 Q6 Relative motion The speed of B as observed from A is given by π£π΅/π΄ = π£π΅ β π£π΄ The easiest way to do this is by breaking it up into x and y components. All the units are in mph, but since we are only dealing with speeds we can leave the units like that throughout. (If we wanted to find distances or momenta, we would need to convert to m/s) x components π£π΄π₯ = π£π΄ cos ππ΄ = 400 cos 45 = 282.8 ππβ π£π΅π₯ = 500 cos 90 =0 or you can just say that vB is entirely vertical, so the horizontal component is zero. y components π£π΄π¦ = π£π΄ sin ππ΄ = 400 sin 45 = 282.8 ππβ π£π΅π¦ = 500 sin 900 = 500 or you can just say that vB is entirely vertical and so equals 500. So we have π£π΅/π΄π₯ = 0 β 282.8 = β282.8 ππβ π£π΅/π΄π¦ = 500 β 282.8 = 217.2 ππβ The total speed is then found by pythagoras |π£π΅ | = β(β282.8)2 + (217.2)2 = 356.6 β 357 ππβ 31 PHY117 Midterm Solutions Q7 D) 2011 Q7 Scaling (with Energy) The rabbit is being scaled by a factor of 2. So length values scale by 2, area values scale by 22 = 4, and volume values scale by 22 = 8. We check each case individually. But first, remember that this is a great place to practice the TRUE/FALSE/MAYBE method for multiple choice we discussed in class! A) The force exerted by a muscle depends on its cross-section area, so this scales by 4. TRUE B) The distances all scale by 2. TRUE C) The gravitational potential energy is given by ππΈπ = ππβ The mass depends on the volume, so it scales by 8. The gravitational acceleration is a constant which doesn't change. But what about the height? We can tell this from conservation of energy. What's more, this is a special case we can remember: this saves us a lot of work if we see the problem again! The gravitational potential energy comes from the initial kinetic energy when the rabbit jumps. The initial kinetic energy comes from the work done by the muscles. The work done by muscles is given by π = πΉ. π We know the force increases by a factor of 4. The distance increases by a factor of 2. So the total work done increases by a factor of 8, and the total energy of the rabbit increases by a factor of 8. We've already seen the mass in the gravitational potential energy term increases by a factor of 8, so the height must remain unchanged. Likewise, if we look at the kinetic energy 1 πΎπΈ = ππ£ 2 2 we see that the mass increases by a factor of 8, so if the total energy also only increases by a factor of 8, the velocity must remain unchanged. C) is TRUE D) We know the mass depends on volume, and increases by a factor of 8, not 4. THIS IS FALSE. E) This is TRUE (see the discussion in section C) 32 PHY117 Midterm Solutions Q8 E) 2011 Q8 Forces There is no friction on the table, so the 5 kg block will fall down and the 15 kg block will be pulled to the right. There are two ways of dealing with this, either looking at the blocks separately or together. Separately Both blocks are moving at the same rate, with the same speed, and size of acceleration a. Looking at the forces on the top block. It will slide to the right with acceleration a. We have Ξ£πΉπ‘ππ π₯ = ππ π = ππ‘ππ π π = 15π Looking at the hanging block, it will fall down with acceleration (-a) (negative because it is going down). 33 PHY117 Midterm Solutions 2011 Q8 Ξ£πΉβπππ π¦ = πβπππ (βπ) +π β πβπππ π = βπβπππ π 15π β 5π = β5π 20π = 5π π= 5(9.8) 20 π = 2.45 π/π 2 and π = 15(2.45) = 36.75 π β 37 π ALTERNATE METHOD: SHORTCUT We can also view the whole system as a single mass moving along a single axis, along the rope. We take the positive direction moving to the right and down along the rope. This gives us Ξ£πΉ = ππ‘ππ‘ππ π +πβπππ π = (ππ‘ππ + πβπππ )π 5(9.8) = (5 + 15)π 5(9.8) = 20π π= 5(9.8) 20 π = 2.45 π/π 2 Then we look at one block, say the top block, and find Ξ£πΉπ‘ππ π₯ = ππ‘ππ π +π = 15(2.45) = 36.75 π β 37 π 34 PHY117 Midterm Solutions Q9 B) 2011 Q9 Forces and Friction Note that we're given the weight in Newtons, so that's already a force. We don't need to multiply by g. We want to move the top five books, so object we're moving has the weight of five books, 5(5) = 25 N This experiences friction along the bottom, between the fifth and bottom books, in the opposite direction to the intended motion. (It would also be possible to start the fifth book only sliding out from between the other books, but that would take more force, and we're asked for the minimum force so we don't have to worry about that.) The minimum force required to start motion in this case is the maximum static friction πΉ = ππ = ππ π We find the Normal force between the fifth and sixth books by looking in the y direction for the five stacked books. They're not accelerating up or down so their acceleration in y is zero. Ξ£πΉπ¦ = 0 πβπ =0 π=π π = 25 And we have πΉ = ππ π = 0.2(25) = 5.0 π 35 PHY117 Midterm Solutions Q10 C) 2011 Q10 Forces and friction along a slope When the mass is sliding we have the kinetic frictional force, so we use ππ = 0.520 The frictional force is given by ππ = ππ π We need the normal force between the slope and the block. We can remember this formula (faster for use in multiple choice questions!) or we can look at the forces. We take tilted axes along the slope. Since the block is not accelerating off or into the slope, in the tilted y direction we have Ξ£πΉπ¦ = 0 π β ππ cos π = 0 π = ππ cos π As you can see, just remembering that formula for a block on a slope with no angled forces (a common situation in 117) is a faster technique. Now we have ππ = ππ ππ cos π = 0.520 (10)(9.8) cos 25 = 46.2 π 36 PHY117 Midterm Solutions Q11 A) 2011 Q11 Forces, friction, and Equations of Motion Work out what you need to do by looking at the keywords in the question. The "minimum stopping distance" means he must come to a stop over a distance. He's changing his velocity from 15.0 m/s to 0 m/s, and we only care about distance, so we use π£ 2 = π£02 + 2πΞπ₯ 0 = 152 + 2πΞπ₯ We have two unknowns, so we need another equation. But before we go, we rearrange to get what we want: 2πΞπ₯ = β152 The minimum stopping distance is then Ξπ₯ = β152 2π Now, we need some other equation. Friction is mentioned, so that is what we use. The maximum force he can exert on the crate is the maximum force of static friction (if he applies a larger force, the crate will slide and the crystal will be in trouble). So the maximum force is πΉ = ππ ππ = ππ π Since the crate is on a level surface, the normal force is the weight, N = mg, and we have ππ = ππ ππ π = ππ π π = 0.4(9.8) = 3.92 π/π 2 This acceleration must be in the negative directino, because the crate is slowing down. π = β3.92 π/π 2 We plug that back into the first equation and we have Ξπ₯ = β152 2(β3.92) = 28.7 π 37 PHY117 Midterm Solutions Q12 B) 2011 Q12 Work and friction The work done is given by π = πΉ. π The force is friction. Normally we find the frictional force from ππ = ππ π but here they haven't given us the coefficient of friction. There must be another way to calculate the force of friction. Well, if they don't give us the friction values, we must look at the force equation. Ξ£πΉ = ππ πΉπππππππ β ππ = ππ 200 β ππ = 55(2) ππ = 200 β 55(2) ππ = 90 π This is in the opposite direction to the 10 m distance travelled, so the dot product is negative, from π. π = ππ cos π So we have π = β90(10) = β900 π½ 38 PHY117 Midterm Solutions Q13 A) 2011 Q13-14 Work The keyword is "constant velocity". Constant velocity means zero change in acceleration, zero change in kinetic energy. And since the cart is moving horizontally, there is no change in potential energy. So there's no change in total energy. And if there's no change in total energy, there is no work done. Q14 C) Conservation of energy Since there is no friction, the total energy of this system is conserved. The total energy at the point where the bolck is released is equal to the energy at x = 0. When the block is released, its speed is zero. When the block is at x =0, the spring potential energy is zero. So we have πΈππππ‘πππ + πππ = πΈπππππ There are no nonconservative forces acting, so πΈππππ‘πππ + 0 = πΈπππππ πΈππππππ π = πΈπ₯=0 ππΈπ ππππππ π + πΎπΈππππππ π = ππΈπ π₯=0 + πΎπΈπ₯=0 ππΈπ ππππππ + 0 = 0 + πΎπΈπ₯=0 1 2 1 ππ₯ = ππ£ 2 2 2 ππ₯ 2 = ππ£ 2 (4 × 102 )(0.05)2 = 5π£ 2 π£ 2 = 0.2 π£ = 0.447 π/π 39 PHY117 Midterm Solutions Q15 E) 2003 Q1 Forces The reading on the scale is the normal force acting on the person from the scale. Looking at the forces acting on the person Ξ£πΉπ¦ = ππ A key point is finding the acceleration. The person is riding down, so the velocity is negative. But they are slowing down, so their acceleration is in the opposite direction, so the acceleration is positive. π β ππ = ππ π β 80(9.8) = 80(2.4) π = 80(9.8) + 80(2.4) = 976 π β 980 π 40 PHY117 Midterm Solutions 2003 Q1 2003 midterm Q1 B) Scaling When we double the linear dimensions, the lengths are multiplied by 2, the areas and multiplied by 4, and the volumes (and so masses) are multiplied by 8. We check each answer. Note that this is a great question for practicing the TRUE/FALSE/MAYBE system we described in class. A) The weight will increase by 8. FALSE B) Why would it need to eat more? Because it's losing more energy in the form of heat through its skin, the surface area, which has increased by a factor of 4. In our simple scaling this looks TRUE. C) The strength of the muscles depends on their cross sectional area, which has increased by 4. The weight depends on the mass, and so the volume, which has increased by 8. So the new strength:weight ratio is 4:8. Which is the same as 1:2. It now has half as much strength (1) as weight (2). The strength:weight ratio hasn't doubled, it has halfed! FALSE D) The surface area has increased by 4. FALSE. E) How likely is it to break bones? The bones' strength depends on their area, which has increased by 4. But the weight they must support has increased by 8, which is twice as much. The animal is now MORE likely to break bones. FALSe. 41 PHY117 Midterm Solutions Q2 A) 2003 Q2 Equations of motion We have two stones, and we want to know when they cross, so we want to know when y1 = y2. The first stone starts from the bottom of the cliff and is thrown up, so y10 = 0 and v10 = +9 m/s The second stone starts at the top of the cliff and is thrown down, so y20 = 6 and v20 = - 9 m/s We use 1 Ξπ¦ = π£0 π‘ + ππ‘ 2 2 1 π¦ β π¦0 = π£0 π‘ β ππ‘ 2 2 π¦ = π¦0 + π£0 π‘ β 4.9 π‘ 2 The stones cross when π¦1 = π¦2 π¦10 + π£10 π‘ β 4.9π‘ 2 = π¦20 + π£20 π‘ β 4.9π‘ 2 π¦10 + π£10 π‘ = π¦20 + π£20 π‘ 0 + 9π‘ = 6 β 9π‘ 18π‘ = 6 π‘ = 0.333 π Now we know when theye cross, we can find where that happens by looking at either stone. Looking at the first π¦1 = π¦0 + π£10 π‘ β 4.9π‘ 2 = 0 + 9(0.333) β 4.9(0.333)2 = 2.46 π 42 PHY117 Midterm Solutions Q3 B) 2003 Q3 Equations of Motion The tile is accelerating linearly due to gravity, with a = - g. The average speed of the tile as it passes the window is found from ππ£πππππ π ππππ = πππ π‘ππππ 1.6 = = 8 π/π π‘πππ 0.2 This is moving downwards, so the velocity is ππ£πππππ π£ππππππ‘π¦ = β8 π/π Since it's accelerating linearly, this is the velocity of the block at the midpoint of the window, 0.8 m below the top of the window. (If it was firing boosters or something we couldn't say this). We know the tile started at a velocity of 0, at rest at the top of the building. So to find the distance between the points (we don't care about time) we can use π£ 2 = π£02 + 2πΞπ¦ (β8)2 = 02 + 2(β9.8)Ξπ¦ β64 = β19.6Ξπ¦ Ξπ¦ = 3.27 π This is the distance from the roof to the midpoint of the window. So the distance from the roof to the top of the window is πππ π‘ππππ = 3.27 β 0.8 = 2.47 β 2.5 π 43 PHY117 Midterm Solutions Q4 A) 2003 Q4-5 Projectile motion To find the angle of impact with the ground, we need the x and y components of the velocity. Then πππππππ‘ = tanβ1 ( π£π¦ ) π£π₯ For projectile motion the x component of the velocity does not change, so the impact x component equals the release x component. Which is π£0π₯ = π£0 cos π = 97.5 cos 50 = 62.7 π/π and so π£π₯ = 62.7 π/π The y component of the release velocity is π£0π¦ = π£0 sin π = 97.5 sin 50 = 62.7 π/π We know how far it falls before hitting the ground (and don't care about time), so we can find y component of the impact velocity with 2 π£π¦2 = π£0π¦ + 2πΞπ¦ BE CAREFUL! Remember that Ξπ¦ = π¦ β π¦0 That's the NEW height minus the OLD height, so we have Ξπ¦ = 0 β 732 = β732 It's an easy mistake to forget the sign! π£π¦2 = 74.72 + 2(β9.8)(β732) = 19927 π/π π£π¦ = β141 π/π NOTE that the square root of a number can be positive or negative. We know this velocity will be negative because the crate is falling down. This makes sense: the package started by going up, because it kept going with the velocity of the plane when it was released, but gravity causes it to fall downward. Now we have 44 PHY117 Midterm Solutions π = tanβ1 ( = tanβ1 (β 2003 Q4-5 π£π¦ ) π£π₯ 141 ) 62.7 = β66.0° The package strikes with a velocity pointing 66.0° below the horizontal. Q5 E) Relative motion (plane) In a plane problem we draw the resultant vector in the direction we want to go. This is due west. Then we draw the wind pointing towards the pointed end of that arrow. This is the wind. This wind is blowing FROM the south, so it is blowing north. We must then fly to the bottom of the wind arrow, as shown We can instantly see we must fly south of west, which cancels out some of the answers. To find the angle, we use π = sinβ1 ( πππππ ππ‘π ) βπ¦πππ‘ππππ’π π = sinβ1 ( π£π€πππ ) π£πππππ = sinβ1 ( 38 ) 245 = 8.92° 8.92° south of west, answer E) 45 PHY117 Midterm Solutions Q6 E) 2003 Q6 Scaling (with the Force of gravity) A classic scaling question The weight of an object is the force of gravity acting on it. This is given by πΉπ = πΊππ π2 If we have ππ΄ = 0.6ππ΅ and πΉπ΄π = πΉπ΅π then we have πΊππ΄ π πΊππ΅ π = ππ΄2 ππ΅2 ππ΄ ππ΅ = 2 ππ΄2 ππ΅ 0.6ππ΅ ππ΅ = 2 ππ΄2 ππ΅ 0.6 1 2 = 2 ππ΄ ππ΅ Rearrange to get our ratio 0.6 = ππ΄2 ππ΅2 ππ΄ = β0.6 ππ΅ = 0.77 46 PHY117 Midterm Solutions Q7 C) 2003 Q7 Forces (elevator) The reading on the scale is the normal force exerted between the scale and the box. This force is given by looking at the forces on the box Ξ£πΉπππ₯ π¦ = ππ π β ππ = ππ π = 60(9.8) + 60π To find the normal force N we need to know the acceleration of the box. This is given by looking at the forces on the entire system, the elevator and scale with box Ξ£πΉπ‘ππ‘ππ π¦ = ππ‘ππ‘ππ π π β ππ‘ππ‘ππ π = (815 + 60)π 9410 β (815 + 60)9.8 = 875π 835 = 875π π = 0.954 π/π 2 and so π = 60(9.8) + 60(0.954) = 645 πππ€π‘πππ 47 PHY117 Midterm Solutions Q8 E) 2003 Q8 Frictional forces This is a problem with two parts. That doesn't make it complicated, we just do one part, then the other. First part: top block only slides To make the top block just begin to move, the total horizontal force is zero (think of it as the force just exactly balancing the maximum friction, so that things are about to move) π΄πΉπ₯ = 0 +πΉπ‘ππ β ππ πππ‘π€πππ = 0 πΉπ‘ππ = ππ πππ‘π€πππ = ππ ππππ‘π€πππ The system is horizontal with no angled forces, so the normal force is just the weight of the top block. ππππ‘π€πππ = ππ πΉπ‘ππ = ππ ππ We are given the applied force, so we know 47 = ππ ππ Now we look at the second situation 48 PHY117 Midterm Solutions 2003 Q8 Second part: bottom block only slides Now we are applying a force to the bottom block, which must equal the maximum static frictional force. This is the total frictional force with the ground AND with the upper block. The coefficient of static friction is the same in each case. π΄πΉπ₯ = 0 +πΉπππ‘π‘ππ β ππ ππππ’ππ β ππ πππ‘π€πππ = 0 πΉπππ‘π‘ππ = ππ ππππ’ππ + ππ πππ‘π€πππ = ππ πππππ’ππ + ππ ππππ‘π€πππ We already know the maximum static friction between the blocks. ππ πππ‘π€πππ = ππ ππ = 47 To find the Normal force with the ground, again, that equals the total weight pressing down from above. This is the weight of two blocks. πππππ’ππ = 2ππ And we have ππ ππππ’ππ = ππ (2ππ) = 2(ππ ππ) = 2(47) = 94 πππ€π‘πππ And we have a total of πΉπππ‘π‘ππ = 94 + 47 = 141 πππ€π‘πππ 49 PHY117 Midterm Solutions Q9 D) 2003 Q9 Forces It's important to include every force acting on the motorcycle. They've given us propulsion and air resistance, but the motorcycle is going up a ramp so there will also be the component of weight acting along the slope, given by πΉππ₯ = ππ sin π = 292(9.8) sin 30 = 1431 πππ€π‘πππ This will be acting down the slope, against the direction of the propulsive force. We have Ξ£πΉπ₯ = ππ πππππ’ππ πππ β πππ πππ ππ π‘ππππ β ππ₯ = ππ 3150 β 250 β 1431 = 292π 1470 = 292π π = 5.03 π/π 2 50 PHY117 Midterm Solutions Q10 D) 2003 Q10 Forces (ramp) The blocks are connected by a string, so they all move the same amount, they all have the same magnitude of velocity and acceleration. We can find this acceleration by looking at each block, or we can use a shortcut by considering all the blocks along the rope as one system. We take the positive direction as up the slope, so that A is moving to the left and B is falling down. Now the force equation for the blocks is Ξ£πΉπππππ ππππ = ππππ ππππππ π +ππ΅ π β ππ΄ π sin π β πππ΄ = (ππ΄ + ππ΅ )π We also have mB = 2mA, so 2ππ΄ π β ππ΄ π sin π β πππ΄ = (ππ΄ + 2ππ΄ )π 2ππ΄ π β ππ΄ π sin π β πππ΄ = 3ππ΄ π We can find the friction acting on block A from πππ΄ = ππ ππ΄ There are no angled forces, so the Normal force with the slope equals the component of the weight into the slope, from Ξ£πΉπ΄π¦ = 0 ππ΄π¦ β ππ΄π¦ = 0 ππ΄π¦ = ππ΄ π cos π so we have πππ΄ = ππ ππ΄ π cos π Plugging that back in, we get 2ππ΄ π β ππ΄ π sin π β ππ ππ΄ π cos π = 3ππ΄ π 2π β π sin π β ππ π cos π = 3π π= π (2 β sin π β ππ cos π) 3 51 PHY117 Midterm Solutions Q11 A) 2003 Q11 Forces and Pythagoras What's happening here? The block has a weight acting down, but is swinging out to the side. The train must be accelerating, providing a force which swings the block out to the side. The block is swinging back to the left, so the train must be accelerating to the right. We can find the values we want from Pythagoras. The total force on the block is the reading on the spring scale, 12 N. The horizontal force would give us the horizontal acceleration, but we can't find that without the angle. We can find the angle from other things we do know. The vertical component is the weight of the block, mg = 1(9.8) = 9.8 N. So the angle can be found from ππ π = cos β1 ( ) π = cos β1 ( 9.8 ) 12 = 35.2° Now that we have the angle, we can find the horizontal component of the force acting on the block πΉπ₯ = 12 sin 35.2 = 6.92 πππ€π‘ππ This is the force swinging the block to the side, so πΉπ₯ = πππ₯ 6.92 = (1)ππ₯ ππ₯ = 6.92 π/π 2 β 7 π/π 2 The block is swinging back to the left, so the train must be accelerating to the right. 52 PREP101 Q12 C) PHYS117 MIDTERM SOLUTIONS 2003 Q12-13 Forces (3rd Law) This is the classic question of Newton's third law. If A exerts a force on B, B exerts the SAME force on A. It doesn't matter what difference in masses exists. So the answer is C or D. The acceleration is given by πΉ = ππ π= πΉ π Since the father has twice the mass, he will have half the resulting acceleration. This gives answer C). Q13 A) Friction If we have increased the angle until the object starts to move, the component of weight along the slope exceeded the maximum static frictional force. The object now experiences kinetic friction. Since the kinetic friction and weight components are constant, the total force is constant, the total acceleration is constant, and the object will continue to speed up at a constant rate. 53 COURSE CRAM Q1 E) Midterm Supplement Solutions Extra Torque Questions Statics, torque and forces First we draw a free body diagram of the ladder. There will be normal forces where it contacts the wall and the floor. There will be weight forces at the center of the ladder, and the location of the man. The frictional force is static, because the point of contact is not slipping, and to the right, to prevent the ladder slipping out to the left. (The only other horizontal force on the ladder is to the left, the normal force with the wall). We always start with torques, and we always take the torques at the point with the most unknowns. Taking the torques around the base of the ladder π΄π = 0 βππ π(2.5 sin 30) β ππΏ π(5 sin 30) + ππ€ (10 sin 60) = 0 β80(9.8)(2.5 sin 30) β 20(9.8)(5 sin 30) + ππ€ (10 sin 60) = 0 and we have ππ€ = 168 After the torques, we always look at the forces π΄πΉπ₯ = 0 ππ β ππ€ = 0 ππ β 168 = 0 ππ = 168 π To two significant figures in scientific notation, 168 β 1.7 × 102 = 170 π COURSE CRAM Q2 C) Midterm Supplement Solutions Extra Torque Questions Torque First we draw a free body diagram. Don't forget that the stick has a 1 N weight, which will be at the midpoint! We take the torque around the point with the most unknowns that we don't want, so we take it around the left end. The torque equation is then π΄π = 0 β2(0.1) β 1(0.5) β 3(0.5) β 3(0.6) + π100 1 = 0 π100 = 4 π BONUS PRACTICE Then if we wanted to find the tension in the other rope Ξ£πΉπ¦ = 0 π0 β 2 β 1 β 3 β 3 + 4 = 0 π0 = 5 π COURSE CRAM Midterm Supplement Solutions Extra Torque Questions Q3 Torque D) When the bucket is "just" raising, the total torque on the crank is zero. Draw the free body diagram of the cran k. π΄π = 0 +ππ β πΉπ = 0 The tension in the cable is provided by the weight of the bucket. Looking at the bucket, which has zero acceleration because the crank is only "just" moving, π΄πΉπ¦ = 0 π β ππ = 0 π = ππ and in the torque equation we have πππ β πΉπ = 0 23(9.8)0.08 β πΉ(0.25) = 0 πΉ = 72.1 π COURSE CRAM Midterm Supplement Solutions Extra Torque Questions Q4 E) Equilibrium means Ξ£F = 0 and Ξ£Ο = 0. Look at each, taking Ο around the center (if itβs in equilibrium, Ξ£Ο will be zero around any point) 1) Ξ£F = -F β 2F + 3F = 0 Ξ£Ο = +F(r) β 2F(r) 2) Ξ£F = +3F β F β 2F = 0 Ξ£Ο = -3F(r) -2F(r) = -5Fr 3) Ξ£F = F β 2F + F = 0 Ξ£Ο = -F(r) + F(r) = 0 Only 3) is in equilibrium