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Unformatted text preview: 1 Problem 1: Hooke’s Law Description: Analyze the force of springs on Haitian taptaps as an application of Hooke’s law. Learning Goal: To understand the use of Hooke’s law for a spring. Hooke’s law states that the restoring force F on a spring when it has been stretched or compressed is proportional to the displacement of the spring from its x equilibrium position. The equilibrium position is the position at which the spring is neither stretched nor compressed. Recall that F ∝ means that F is equal to a constant times . For a spring, the proportionality constant x x is called the spring constant and denoted by k. The spring constant is a property of the spring and must be measured experimentally. The larger the value of k, the stiﬀer the spring. In equation form, Hooke’s law can be written F = −k . x The minus sign indicates that the force is in the opposite direction to that of the spring’s displacement from its equilibrium length and is "trying" to restore the spring to its equilibrium position. The magnitude of the force is given byF = kx, where x is the magnitude of the displacement. In Haiti, public transportation is often by taptaps, small pickup trucks with seats along the sides of the pickup bed and railings to which passengers can hang on. Typically they carry two dozen or more passengers plus an assortment of chickens, goats, luggage, etc. Putting this much into the back of a pickup truck puts quite a large load on the truck springs. A truck has springs for each wheel, but for simplicity assume that the individual springs can be treated as one spring with a spring constant that includes the eﬀect of all the springs. Also for simplicity, assume that all four springs compress equally when weight is added to the truck and that the equilibrium length of the springs is the length they have when they support the load of an empty truck. Part A A 60 − kg driver gets into an empty taptap to start the day’s work. The springs compress0.02m. What is the eﬀective spring constant of the spring system in the taptap? Hint A.1 How to approach the problem: The compression of the springs is governed by Hooke’s law. The amount the springs are compressed when the driver climbs into the truck is given in the problem statement. The force that acts to compress the springs is the force caused by the driver getting into the truck. F = −k x k= solve for k
mag (F ) mag ( ) x k = 29000 ANSWER: 2.9 × 104 N/m 1.1 Part B After driving a portion of the route, the taptap is fully loaded with a total of 25 people with an average mass of 60kg per person. In addition, there are three 15 − kg goats, ﬁve 3 − kg chickens, and a total of 25kg of bananas on their way to the market. Assume that the springs have somehow not yet compressed to their maximum amount. How much are the springs compressed? Hint B.1 How to ﬁnd the compression of the spring: The spring compression is governed by Hooke’s law. Use the spring constant you found in Part A. To ﬁnd the force add the total weight of the load on the truck. 1 so if we solve for x we have: = x F/k F= F = ( m) ∗ g = ANSWER: 0.54m from above we have:F = −k x (25 people × 60 kg/person + 3 goats × 15 kg/goat + 5 chickens × 3 kg/chickens + 25 kg bananas) ∗ g ⇒ x = 1585 ∗ 9.8/2900 = 0.54 1.2 Part C Whenever you work a physics problem you should get into the habit of thinking about whether the answer is physically realistic. Think about how far oﬀ the ground a typical small truck is. Is the answer to Part B physically realistic? Select the best choice below. ANSWER: No, typical small pickup truck springs are not large enough to compress about a half meter. The answer to Part B is not physically realistic because the springs of a typical light truck will compress their maximum amount (typically about 10 \rm cm) before the total weight of all the passengers and other cargo given in Part B is added to the truck. When this maximum compression is reached, the springs will bottom out, and the ride will be very rough. 1.3 Part D Now imagine that you are a Haitian taptap driver and want a more comfortable ride. You decide to replace the springs with new springs that can handle the typical heavy load on your vehicle. What spring constant do you want your new spring system to have? ANSWER: The new springs should have a spring constant that is SUBSTANTIALLY LARGER than the spring constant of the old springs. A spring constant with a large value is a stiﬀ spring. It will take more force to compress (or stretch) a stiﬀ spring. On a taptap, stiﬀer springs are less likely to bottom out under a heavy load. However, with a lighter load, for most vehicles, very stiﬀ springs will not compress as much for a bump in the road. Hence very stiﬀ springs will give a better ride with a very heavy load, but lessstiﬀ springs (lower spring constant) will give a smoother ride with a light load. This is why larger vehicles need stiﬀer springs than smaller vehicles. 2 Problem 5.1 Description: A horizontal force accelerates a box across a rough horizontal ﬂoor with friction present, producing an acceleration. If the force is now tripled, but all other conditions remain the same, the acceleration will become a=
F − Ff r m (2.1) so if we have a force of 3F then we have an acceleration equal to 2 a= 3F − F f r m (2.2) but the problem asks us to compare (2.2) to 3a. So if we multiply (2.1) by three we get:
fr 3a = which is clearly less than(2.2), which makes intuitive sense because if we increase the force, we m aren’t increasing the friction. We would have to increase the weight, and thus the normal force to increase the force of friction. ANSWER: greater than 3a 3 F − 3F 3 Problem 5.2 A child of mass M using a parachute reaches a terminal velocity V, with the air resistance directly proportional to the square of the child’s velocity. If an adult of mass 3M uses the same parachute under exactly the same conditions, his terminal velocity will be closest to: If we draw a free body diagram for this problem we get this equation: FD = Dv 2 − mv Since we’re looking at TERMINAL velocity, we can set FD to zero and solve for velocity which gives us: v = mg D so if D is the same, then we can compare two falling humans of diﬀerent mass. v1 = m1 g and v2 = m2 g D D Since m2 = 3m1 √ √ v2 = 3m1 g = 3 m1 g = 3v1 D D ANSWER: √ V3 4 Problem 5.3 An artist wearing spiked shoes pushes two crates across her frictionless horizontal studio ﬂoor. (See the ﬁgure below.) Part A If she exerts a horizontal 56.0 N force on the smaller crate, then the smaller crate exerts a force on the larger crate that is closest to... The Freebody Diagram for the smaller Crate:
F21 ←−−→56 →ma The Freebody Diagram for the smaller Crate:
−→F12 →ma m1 a = 56 − F21 (4.1) m2 a = F12 (4.2) We know two things about this problem: 1) F12 = F21 and 2) the acceleration is the same for both blocks, so if we solve for a in (4.1) and F12 in (4.1) then you can solve for a, like this: a= a=
56−m2 a m1 56 m m1 (1+ m2 )
1 =⇒ a= 56 m1 if we plug in m1 and m2 , we get a = 2.33. − m2 m1 a =⇒ a(1 + m2 m1 ) = 56 m1 plugging a into (4.2) we get F12 = F21 = 46.7 ANSWER: greater than 46.7N 3 5 Multiple Choice Problem 5.5 Description: A horizontal pull P drags two boxes connected by a horizontal rope having tension T, as shown in the ﬁgure . The ﬂoor is horizontal and frictionless. Part A Decide which of the following statements is true without doing any calculations: The key to this answer is to notice that the force, P , is pulling two boxes totalling 375 lb. The force, T , is only pulling one box of 350 lb. So by inspection, its clear that the force, P , is pulling the ﬁrst box and the tension pulling against it. ANSWER: P >T 6 Multiple Choice Problem 5.8 Description: A weightless spring scale is attached to two equal weights as shown in the ﬁgure. Part A The reading in the scale will be The key to this problem is to imagine that one of the weights is replaced by a wall, such that the scale is attached to string attached to a wall on one side, and a weight on the other. If this were the case, it would be clear that only one weight would be pulling on the scale and it would be reading a weight of W. Conceptually, these two problems are the exact same. The scale is in equilibrium, one weight counterbalanced by the other, hence the answer is the same. ANSWER: W 7 Multiple Choice Problem 5.9 Description: Two objects are connected by a light wire as shown in the ﬁgure, with the wire pulling horizontally on the 400N object. Part A After this system is released from rest, the tension in the wire will be... By adding up all the forces in the x and y directions for both masses, m1 on the table and m2 hanging oﬀ the table. Because m1 is only moving in the x direction we get: m1 a1 = T − µN1 = T − µm1 g and for m2 : m2 a2 = m2 g − T Because both m1 and m2 are attached to the same wire, they are traveling at the same speed hence a1 = a2 setting the equations equal we can solve for T. T = 1/2 ∗ (m1 + µm2 )g 4 This problem oﬀers two coeﬃcients of friction, µstatic and µkinetic . The key to choosing which coeﬃcient to use is noting that the system is no longer at rest which means that it is moving. So the proper coeﬃcient to choose is µkinetic . ANSWER: 200N 8 Problem 5.9 Traction apparatus. Description: In order to prevent muscle contraction from misaligning bones during healing (which can cause a permanent limp), injured or broken legs must be supported horizontally and at the same time kept under tension (traction) directed along the leg. One version of a device to accomplish this aim, the Russell traction apparatus, is shown in the ﬁgure. This system allows the apparatus to support the full weight of the injured leg and at the same time provide the traction along the leg. 8.1 Part A If the leg to be supported weighs 41.0 N, what must be the weight of W? By inspecting the problem, its clear that the only force in the ydirection, supporting the leg, is that of the weight. If we drew a free body diagram for each pully, summed up the forces, and solved for tension in the y direction we would ﬁnd that the tension in the wire supporting the leg has to be equal to the wieght of the leg. ANSWER: Wleg = 41.0N 8.2 Part B What traction force does this system produce along the leg? Fx = Ftraction = T1x + T2x = (T1 + T2 )cos25o T1 = T2 = W Ftraction = 2 ∗ W ∗ cos25o ANSWER: F = 1.81262 ∗ wleg = 74.3N 9 Problem 5.11 Description: Two artifacts in a museum display are hung from vertical walls by very light wires, as shown in the ﬁgure. The most direct way to solve this problem is to make a free body diagram. From past examples, we know that a mass hanging from a wire is going to produce a tension that is equal and opposite to the force of 5 gravity on the object hanging. So if we take a free body diagram and sum the forces for the ball, denoting T1 as the horizontal wire, and T2 as the inclined wire we get: Fx = T1 − T2 sin53o = 0 Fx = T2 cos53o − m1 g − m2 g = 0 9.1 Part A Find the tension in the horizontal wire. To do this we ﬁrst need to ﬁnd the tension in the inclined wire. Solving for T2 above gives us: T2 = (m1 g + 175)/cos53o = 812N Now that we have T2 we can solve for T1 : T1 = T2 sin53o ANSWER: T1 = 648N 9.2 Part B Find the tension in the inclined wire. as we found from above. ANSWER: T2 = 812N 9.3 Part C Find the tension in the vertical wire. As above, T3 = mg ANSWER: T3 = 314N 9.4 Part D Would the answers be diﬀerent if each wire were twice as long, but the angles were unchanged? None of the tensions rely on length, so the answer is: ANSWER: NO 10 Problem 5.22 Description: A short train (an engine plus four cars) is accelerating at 1.10m/s2 . The mass of each car is 38000kg, and each car has negligible frictional forces acting on it. In solving this problem, note the importance of selecting the correct set of cars to isolate as your object. 6 10.1 Part A What is the force of the engine on the ﬁrst car? ANSWER: By knowing that Fx = ma, and for each car, ma = Tf ront − Tback which is basically saying that the net acceleration is proportional to Tension pulling the car minus the Tension its pulling. So for the car one we’ve got Fx = max = 4 ∗ ma = 4 ∗ 38000kg ∗ 1.10m/s2 ANSWER: F1 = 1.67 ∗ 105 N 10.2 Part B What is the force of the ﬁrst car on the second car? This time we only have 3 cars so: Fx = max = 3 ∗ ma = 3 ∗ 38000kg ∗ 1.10m/s2 ANSWER: F2 = 1.25 ∗ 105 N 10.3 Part C What is the force of the second car on the third car? two cars: Fx = max = 2 ∗ ma = 2 ∗ 38000kg ∗ 1.10m/s2 ANSWER: F3 = 8.36 ∗ 104 N 10.4 Part D And what is the force of the third car on the fourth car? and ﬁnally: Fx = max = ma = 38000kg ∗ 1.10m/s2 ANSWER: F4 = 4.18 ∗ 104 N 11 Problem 5.28 Description: At a construction site, a 22.0kg bucket of concrete is connected over a very light frictionless pulley to a 375N box on the roof of a building. (See the ﬁgure.) There is no appreciable friction on the box, since it is on roller bearings. The box starts from rest. 7 11.1 Part A Find the acceleration of the bucket. This is very similar to problem 7, above. As before: By adding up all the forces in the x and y directions for both masses, m1 on the table and m2 hanging oﬀ the table. Because m1 is only moving in the x direction we get: m1 a1 = T − µN1 = T − µm1 g because there is no friction, m1 a1 = T and for m2 : m2 a2 = m2 g − T (11.2) Because both m1 and m2 are attached to the same wire, they are traveling at the same speed hence a1 = a2 . so if we plug (11.1) into (11.2) we get: m2 a + m1 a = m2 g − 375 then: a=
m2 g m1 +m2 (11.1) the trick then is to ﬁnd m1 , it which is 375N/9.8m/s2 ANSWER: a = 3.58m/s2 11.2 Part B How fast is the bucket moving after it has fallen 1.50m (assuming that the box has not yet reached the edge of the roof)? so from kinematics we have: y = 1/2at2 so t = 2y/a and v = at so v = a 2y/a ANSWER: a = 3.28m/s2 12 Problem 5.34 Description: Two crates connected by a rope of negligible mass lie on a horizontal surface. (See the ﬁgure.) Crate A has mass mA and crate B has mass mB . The coeﬃcient of kinetic friction between each crate and → − the surface is µk . The crates are pulled to the right at a constant velocity of 3.2m/s by a horizontal force F . 12.1 Part A → − In terms of mA , mB , and µk , calculate the magnitude of the force F . If we look at the whole thing as a system, we can solve this quickly: → − F = ma = 0 = N1 + N2 µk mA g + µk mB g = µk (mA + mB )g ANSWER: F = µk ∗ (mA + mB ) ∗ g 8 12.2 Part B In terms of mA , mB , and µk , calculate the tension in the rope connecting the blocks. If we isolate just block A we have T − µk ma g = 0 ANSWER: T = µ k ∗ mA ∗ g 13 Problem 5.48 Description: A person pushes on a stationary 125N box with 75N at 30◦ below the horizontal, as shown in the ﬁgure. The coeﬃcient of static friction between the box and the horizontal ﬂoor is 0.80. 13.1 Part A What is the normal force on the box? To ﬁnd the normal force, you must isolate the forces in the ydirection. Fy = 0 = mg − N + Fy N = mg + F sin30◦ ANSWER: N = 163N 13.2 Part B What is the friction force on the box? The ﬁrst inclination in this case is to assume that Ff r = µN . in this case, though µN = 130N , and this exceeds the force we’re applying on it in the xdirection, so as the problem states, the box is not moving and we have to check to see if the friction force is equal and opposite the force in the xdirection. Fx = 0 = F x − F f r Ff r = F cos30◦ ANSWER: F = 65.0N 9 13.3 Part C The person now replaces his push with a 75N pull at 30◦ above the horizontal. Find the normal force on the box in this case. N = mg − F sin30◦ ANSWER: N = 87.5N 14 Problem 5.87 Description: A block is placed against the vertical front of a cart as shown in the ﬁgure. (a) What acceleration must the cart have in order that block A does not fall? The coeﬃcient of static friction between the block and the cart is µs . A block is placed against the vertical front of a cart as shown in the ﬁgure. Part A What acceleration must the cart have in order that block A does not fall? The coeﬃcient of static friction between the block and the cart is µs . If we take a free body diagram of the block, the only force acting on it in the xdirection is the force of the cart on the block: Fx = ma = Fcb = N Fy = ma = mg − Ff r = 0 mg = Ff r = µs N ANSWER: a = g/µs therefore: µs ∗ ma = mg 10 ...
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This note was uploaded on 01/17/2011 for the course PHYS 6a taught by Professor Stanek during the Fall '07 term at UCSB.
 Fall '07
 STANEK
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