Force Calculations in Pulley Systems and Shot Put Dynamics
School
Socastee High**We aren't endorsed by this school
Course
MYRTLE BEACH 33243
Subject
Physics
Date
Dec 11, 2024
Pages
10
Uploaded by RyleighCarolineWilliams
Practice Problems βSecond Exam Topics Solutions 1.The figure below shows a man sitting in a bosunβschair that dangles from a massless rope, which runs over a massless, frictionless pulley and back down to the manβs hand. The combined mass of the man and chair is 95.0 π?. With what force magnitude must the man pull on the rope if he is to rise a) with a constant velocity and b) with an upward acceleration of 1.30 π/?2? If the rope on the right extends to the ground and is pulled by a co-worker, with what force magnitude must the co-worker pull for the man to rise c) with a constant velocity and d) with an upward acceleration of 1.30 π/?2? What is the magnitude of the force on the ceiling from the pulley system in e) part a, f) part b, g) part c, and h) part d? a)In this problem setup, the man is pulling down on the rope on the right with a force causing a tension π. If we consider all the forces in the vertical direction acting on the person, we see that there is the tension in the rope connected to the chair and the tension in the rope held by the person in addition to the weight of the person-chair. Since the pulley is massless and frictionless, then the tensions on either side of the pulley are the same. Thus, we can write: Ξ£? = π1+ π2β ? = ππT1= T2= Tβ 2π β π? = ππ = 0β π =12π? = (12) (95)(9.8) = 465.5 ?b)From earlier, we know that: π =12(ππ + π?) =π2(π + ?) = (952) (1.3 + 9.8) = 527.25 ?c)This time, if the rope is pulled by someone else and not the worker in the chair, the net force on the person in the chair is now: π β π? = ππβ π = π(π + ?) = (95)(0 + 9.8) = 931 ?d)π = π(π + ?) = (95)(1.3 + 9.8) = 1054.5 ?e)At the ceiling, the ropes are pulling down with a force of 2π. Thus, we get: 2π = (2) (12π?) = π? = 931 ?
f)2π = (2) (?2(π + ?)) = π(π + ?) = (95)(1.3 + 9.8) = 1054.5 ?g)2π = (2)(π(π + ?)) = 2π(π + ?) = (2)(95)(0 + 9.8) = 1862 ?h)2π = (2)(π(π + ?)) = 2π(π + ?) = (2)(95)(1.3 + 9.8) = 2109 ?2.A shot putter launches a 7.260 π?shot by pushing it along a straight line of length 1.650 πand at an angle of 34.10Β°from the horizontal, accelerating the shot to the launch speed from its initial speed of 2.500 π/?. The shot leaves the hand at a height of 2.110 πand at an angle of 34.10Β°, and it lands at a horizontal distance of 15.90 π. What is the magnitude of the athleteβs average force on the shot during the acceleration phase? Treat the motion during the acceleration phase as though it were along a ramp at the given angle. To find the force on the shot, we need to determine the acceleration of the shot as it traveled the 1.650 π. To determine that, we need to know how fast the shot was moving as it left the shot putterβs hand at the end of the acceleration. To determine how fast the shot was moving, we need to figure out how long it took the shot to travel the horizontal distance of 15.9 π. This will give us the speed because since there is no acceleration in the horizontal direction once the shot leaves the hand, then the time it takes to travel a distance is simply Ξ?Ξ?. To find the time, we can make use of the given height information of the ball and that the free-fall acceleration due to gravity is ?. Ξ? = ?0?? +12π??2Unfortunately, we are looking for the initial velocity or any of its components so we are stuck here. However, we can relate the initial velocity components together. ?0?= ?0sin π?0?= ?0cosπ =??Where ?is the horizontal distance traveled after leaving the shot putterβs hand. Relating this together, we get: ?0?= ?0cosπ =??β ?0=?? cosπ?0?= ?0sin π = (?? cos π) sin π =??tan πβ Ξ? = ?0?? +12π??2= (??tan π)? +12π??2= ? tan π +12π??2β ? = β2(Ξ? β ? tan π)π?
Thus, we get an initial velocity of: ?0=?? cos π= (?cos π)(1β2(Ξ? β ? tan π)π?)= (1cos π)β?2π?2(Ξ? β ? tan π)This initial velocity after leaving the hand is the final velocity at the end of the push from the hand. Thus, we can find the acceleration as: ??2= ?02+ 2πΞ?β π =??2β ?022Ξ?Here, Ξ?is the length traveled during the push, which was 1.650 πand the initial velocity was that given in the problem as 2.5 π/?. Thus, we have: π =((1cos π) β?2π?2(Ξ? β ? tan π))2β ?022Ξ?Now, to find the force the shot putter applied to the shot, we must realize that this acceleration is the acceleration due to the net force acting on the shot in the direction of motion. Working out Newtonβs Second Law equation for along the path traveled, we get: ? β ??= ππ?Here, I am saying that the direction of the path taken while being pushed is the x-axis and the acceleration we determined earlier is that acceleration in the x-direction. Setting up our free body diagram, we see that ??= ? sin π = π? sinπ. Thus, we get: ? β π? sin π = ππ?β ? = ππ?+ π? sinπ = π(π?+ ? sin π) = π(((1cos π) β?2π?2(Ξ? β ? tan π))2β ?022Ξ?+ ? sinπ)= 334.8394 ?3.The figure below shows a box of dirty money (mass π1= 3.0 π?) on a frictionless plane inclined at angle π1= 30Β°. The box is connected via a cord of negligible mass to a box of laundered
money (mass π2= 2.0 π?) on a frictionless plane inclined at angle π2= 60Β°. The pulley is frictionless and has negligible mass. What is the tension in the cord? Since the pulley is frictionless and massless, we know that the tensions on either side of the pulley are the same. Treating each side of the pulley individually and the x-direction as along the slopes for each side with positive going to the right and up (for the left) or down (for the right) the slopes, we get the following net force equations: Ξ£?1?= π1β ?1?= π1π?Ξ£?2?= ?2?β π2= π2π??1?= ?1sinπ1= π1? sin π1?2?= ?2sin π2= π2? sin π2Since we want to solve for the tensions, I will solve for the acceleration in one net force equation and substitute that into the other while also taking into account π1= π2= π. π?= (1π1)(π β π1? sin π1)β π2? sin π2β π = π2((1π1) (π β π1? sin π1)) =π2π1(π β π1? sinπ1)Solving for π, we get: π2? sinπ2β π =π2π1π β π2? sinπ1β (1 +π2π1)π = π2? sin π2+ π2? sinπ1β π =π2?(sinπ1+ sin π2)1 +π2π1=(2)(9.8)(sin 30 + sin60)1 +23= 16.064 ?4.In the figure below, a slab of mass π1= 40 π?rests on a frictionless floor, and a block of mass π2= 10 π?rests on top of the slab. Between block and slab, the coefficient of static friction is π?= 0.60, and the coefficient of kinetic friction is π?= 0.40. A horizontal force ?βof magnitude
100 ?begins to pull directly on the block, as shown. In unit-vector notation, what are the resulting accelerations of a) the block and b) the slab? First we need to see if π2is going to start moving from this applied force and slide across the top of π1. For that, we need to see if the maximum static friction force present is greater than the applied force. Setting to the right as positive, we get: ??,???β ? = π???2β ?Since the smaller block is on a flat and level surface of the slab, we do not need to consider any angles and the normal force the smaller block feels is just its weight. Thus, we get: π???2β ? = π?π2? β ? = (0.6)(10)(9.8) β 100 = β41.2 ?Because the answer is negative, this means that the force applied to π2is going to make it start moving with respect to the slab. This also means that the acceleration of the slab will be different from the block. a)For the block, we write the net force equation with the knowledge that the block will move from this applied force. Ξ£? = ??β ? = π2π2β π2= (1π2) (??β ?) = (1π2) (π?π2? β ?) = (110) ((0.40)(10)(9.8) β 100) = β6.08Thus, the acceleration of the block is πβ = (β6.08 π/?2)?Μb)The force on the slab is what is felt from the block being dragged across the top of the slab. Thus, what the slab feels is the friction force of the block. However, we must realize that the direction is to the left because while the block feels a force from friction to the right, Newtonβs Third Law tells us that the slab will feel the same force to the left. β??2= π1π1β π1= β (1π1) ??2= β??2π1= βπ?π2?π1= β0.98Thus, the acceleration of the slab is πβ = (β0.98 π/?2)?Μ5.In the figure below, a box of ant aunt (total mass π1= 1.65 π?) and a box of ant uncles (total mass π2= 3.30 π?) slide down an inclined plane while attached by a massless rod parallel to the plane. The angle of incline is π = 30Β°. The coefficient of kinetic friction between the aunt box and the incline is π?1= 0.226; that between the uncle box and the incline is π?2= 0.113. Compute a) the tension in the rod and b) the magnitude of the common acceleration of the two boxes. c) How would the answers to a) and b) change if the uncles trailed the aunts?
a)Since the boxes are connected via a rod, they will have the same acceleration. Drawing our free body diagrams where down the slope is the positive x-direction, we get the following net force equations: Ξ£??1= ?1?β ??1+ π1= π1πΞ£??2= ?2?β ??2β π2= π2πHere, the force from the rod on π1is positive, meaning we assumed that the rod is pulling π1. From this, that means that the rod must be pulling back on π2, meaning that the force is then up the slope and thus negative. We also must take care in our assumption, that π2is pulling π1. To check if this assumption is correct, we can consider each box individually and unconnected and then compare their accelerations. That is, if π2has a larger acceleration than π1, then π2is indeed pulling π1down the incline. Ξ£??1= ?1?β ??1= π1π1Ξ£??2= ?2?β ??2= π2π2Here, ?1?= ?1sinπ = π1? sinπ?1?= ?1cos π = π1? cos π?2?= ?2sin π = π2? sin π?2?= ?2cos π = π2? cosπΞ£??1= ??1β ?1?= 0β ??1= π1? cos πΞ£??2= ??2β ?2?= 0β FN2= m2? cos πβ π1=?1?β ??1π1=π1? sinπ β π?π1? cos ππ1= ? sin π β π?1? cos πβ π2=?2?β ??2π2=π2? sinπ β π?π2? cos ππ2= ? sin π β π?2? cos πBecause the coefficient of kinetic friction for π2is smaller than for π1, π2must thus be accelerating faster than π1and therefore pulls π1down the slope so we use our original equations for the net force above. Ξ£??1= ?1?β ??1+ π1= π1πΞ£??2= ?2?β ??2β π2= π2πβ π =π1? sinπ β π?1π1? cos π + ππ1β π2? sin π β π?2π2? cos π β π = (π2π1) (π1? sinπ β π?1π1? cos π + π)
β (1 +π2π1)π = π2? sin π β π?2π2? cos π β π2? sin π + π?1π2? cosπ= π2? cos π (π?1β π?2)β π =π2? cosπ (π?1β π?2)1 +π2π1=(3.3)(9.8) (β32) (0.226 β 0.113)1 +3.31.65= 1.0549 ?b)π =?1? sin πβππ1?1? cos π+π?1= ? sin π β π?1? cos π +?2? cos π(ππ1βππ2)?1+?2= 3.6324 π/?2c)By switching the boxes, the smaller box would now be pushed down the incline instead of pulled. In this situation, the tension in the rod would change directions but remain the same magnitude. The acceleration would remain the same. 6.A police officer in hot pursuit drives her car through a circular turn of radius 300 πwith a constant speed of 80.0 ππ/β. Her mass is 55.0 π?. What are a) the magnitude and b) the angle (relative to the vertical) of the net force of the officer on the car seat? Be sure to consider both horizontal and vertical forces. a)First, we convert the speed to meters per second. 80 ππ/β= 22.2222 π/?The horizontal force acting on the officer, likely friction, must be equal to the centripetal force as she is moving in a circle that is flat with the ground. In the vertical direction, the force of the officerβs weight is pushing down on the car seat. Thus, we can write the net force as:Ξ£? = β??2+ ??2=β(π?2?)2+ (π?)2=β((55)(22.2222)2300)2+ ((55)(9.8))2= 546.5506 ?b)The angle from the vertical can be determined as: π = tanβ1????= tanβ1π?2?π?= tanβ1?2??= tanβ122.22222(300)(9.8)= 9.5349Β°7.A roller-coaster car at an amusement park has a mass of 1200 π?when fully loaded with passengers. As the car passes over the top of a circular hill of radius 18 π, assume that its speed is not changing. At the top of the hill, what are the a) magnitude ??and b) direction (up or down) of the normal force on the car from the track if the carβs speed is ? = 11 π/?? What are c) ??and d) the direction if ? = 14 π/?? Drawing the free body diagram and taking toward the center of the circle as positive, we get the following: a)Ξ£? = ? β ??=??2?β ??= ? βπ?2?= π? βπ?2?= π (? β?2?) = (1200) (9.8 β11218) = 3693.33 ?
b)The way we wrote our net force equation assumes that the normal force is going up. Since our answer was positive, this indicates that the normal force is indeed up. c)? β ??=??2?β ??= π? βπ?2?= π (? β?2?) = (1200) (9.8 β14218) = β1306.67 ?d)Because as before we assumed that the normal force was up but our answer for c) was negative, this indicates that the normal force is actually down. 8.The radius π
βand mass ?βof a black hole are related by π
β= 2??β/?2, where ?is the speed of light. Assume that the gravitational acceleration π?of an object at a distance ?π= 1.001π
βfrom the center of a black hole can be derived from ? =πΊ?1?2?2. a) Derive the general expression for π?. b) In terms of ?β, find π?at ?π. c) Does π?at ?πincrease or decrease as ?βincreases? d) What is π?at ?πfor a very large black hole whose mass is 1.55 Γ 1012times the solar mass of 1.99 Γ 1030π?? a)For an object that is under the influence of gravity and there are no other forces present, we get: ? =??βππ?π2= πππ?β π?=??β?π2b)Here, π
β=2πΊ?β?2, thus we get for the point ?π= 1.001π
β: π?=??β?π2=??β(1.001π
β)2=??β(1.001 (2??β?2))2=?4(2.002)2??β=(2.998 Γ 108)4(2.002)2(6.67 Γ 10β11)?β=3.0218 Γ 1043?βπ? β π/?2c)Because the mass term is in the denominator, π?actually decreases as the mass increases. d)π?=3.0218Γ1043?β=3.0218Γ1043(1.55Γ1012)(1.99Γ1030)= 9.7969 π/?29.The fastest possible rate of rotation of a planet is that for which the gravitational force on material at the equator just barely provides the centripetal force needed for the rotation. a) Why? b) Show that the corresponding shortest period of rotation is π = β3?πΊ?where ?is the uniform density of the spherical planet. c) Calculate the rotation period assuming a density of 3.0 ?/?π3, typical of many planets, satellites, and asteroids. No astronomical object has ever been found to be spinning with a period shorter than that determined by this analysis. a)If the rotation rate were any greater, then loose objects would fly off into space. b)Density is defined as the mass divided by the volume. Thus, we get ? = ?? =43?π
3?. The shortest period of rotation is then the centripetal force equivalent to the gravitational force as:
??ππ
2=π?2π
= ππ
π2=4?2ππ
π2β π2=4?2π
3??=4?2π
3? (43?π
3?)=3???β π = β3???c)? = 3.0 ?/?π3= 3000 π?/π3β π = β3???= β3?(6.67 Γ 10β11)(3000)= 6862.9687 ? β 1.9 βπ???10.In the image above, there are four buckets sitting on top of a table connected to ropes that pass over frictionless and massless pulleys connected to the ceiling. Each bucket is directly below one of each of the pulleys connected to the ceiling. The ropes are then connected to the center of the table thus supporting the table above the ground and make an angle of 30Β°with respect to the horizontal and level table. If each bucket has a mass of 11 π?, what is the maximum possible mass of the table such that the table does not move downward? The table is perfectly square and each bucket is at one corner of the table. For this problem, we have four buckets connected to ropes to support the table they are all sitting upon. Each bucket provides a tension that supplies part of the total force keeping the table in the air. Furthermore, because we are told that the table is perfectly square, that each bucket is at one of the corners of the table, and that each bucket is the same mass, we can conclude that each bucket supplies the same tension to the table in the center. Setting up our net force expressions and setting the buckets moving downward as the positive direction, we get the following:
Ξ£? = π1+ π2+ π3+ π4β ??????= π?????πHere, this is the expression for the net force acting on the table. Because we want the maximum mass possible that the table could have such that it does not move downward, we know that the forces must be balanced and thus we have equilibrium. Furthermore, as justified earlier, each of these tensions are going to be the same. Thus, we get: β Ξ£? = 4π β π?????? = 0Since we are looking for the mass of the table, we must first determine the tension in the ropes. To do this, we can consider the entire system together. That is, we know the tension in the ropes is what is holding up the buckets and table together, so let us consider that. Because the pulleys are frictionless and massless, the tension on either side of each pulley is the same. These tensions are what are keeping the buckets and table elevated. Setting up the net force expression for this system and taking down as negative, we get: Ξ£? = 8π β ??????β 4???????= 0This expression is only valid for when the table and buckets are in contact with each other. Luckily, this is our situation. There is a factor of eight in front of the tension because for each pulley, two ropes are connected to the bucket/table system. β π =18(??????+ 4???????) =18(π?+ 4π?)?Now, we can plug this expression into what we found earlier and solve for the mass of the table. Ξ£? = 4π β π?? = 0β 4 (18(π?+ 4π?)?) β π?? = 0β π?+ 4π?β 2π?= 0β π?= 4π?= 4(11) = 44 π?Thus the table can have a mass of at most 44 π?and not move downward.