This preview shows pages 1–3. Sign up to view the full content.
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
Unformatted text preview: Chapter 5 Applications of Newtons Laws CHAPTER 5 Applications of Newtons Laws 1* Various objects lie on the floor of a truck moving along a horizontal road. If the truck accelerates, what force acts on the objects to cause them to accelerate? Force of friction between the objects and the floor of the truck. 5* A block of mass m is at rest on a plane inclined at angle of 30 o with the horizontal, as in Figure 538. Which of the following statements about the force of static friction is true? ( a ) f s > mg ( b ) f s > mg cos 30 o ( c ) f s = mg cos 30 o ( d ) f s = mg sin 30 o ( e ) None of these statements is true. ( d ) f s must equal in magnitude the component of the weight along the plane. 9* A block of mass m is pulled at constant velocity across a horizontal surface by a string as in Figure 539. The magnitude of the frictional force is ( a ) k mg . ( b ) T cos . ( c ) k ( T mg ). ( d ) k T sin . ( e ) k ( mg + T sin ). ( b ) The net force is zero. 13* The force that accelerates a car along a flat road is the frictional force exerted by the road on the cars tires. ( a ) Explain why the acceleration can be greater when the wheels do not spin. ( b ) If a car is to accelerate from 0 to 90 km/h in 12 s at constant acceleration, what is the minimum coefficient of friction needed between the road and tires? Assume that half the weight of the car is supported by the drive wheels. ( a ) s > k ; therefore f is greater if the wheels do not spin. ( b ) 1. Draw the freebody diagram; the normal force on each pair of wheels is 1/2 mg. 2. Apply F = m a 3. Solve for a 4. Find s f s = ma = s F n ; F n = 1/2 mg a = 1/2 s g = (25 m/s 2 )/(12 s) = 2.08 m/s 2 s = (2 2.08/9.81) = 0.425 17* A 50kg box that is resting on a level floor must be moved. The coefficient of static friction between the box Chapter 5 Applications of Newtons Laws and the floor is 0.6. One way to move the box is to push down on it at an angle with the horizontal. Another method is to pull up on the box at an angle with the horizontal. ( a ) Explain why one method is better than the other. ( b ) Calculate the force necessary to move the box by each method if = 30 o and compare the answers with the result when = 0 o . The freebody diagram for both cases, > 0 and < 0, is shown. ( a ) > 0 is preferable; it reduces F n and therefore f s . ( b ) 1. Use F = m a to determine F n 2. f s,max = s F n 3. To move the box, F x = F cos f s,max 4. Find F for m = 50 kg, s = 0.6, and = 30 o , = 30 o , and = 0 o F sin + F n mg = 0. F n = mg  F sin f s,max = s ( mg  F sin ) F = s ( mg  F sin )/cos ; F = sin cos s + g m s F (30 o ) = 252 N, F (30 o ) = 520 N, F (0 o ) = 294 N 21* Returning to Figure 541, this time m 1 = 4 kg. The coefficient of static friction between the block and the incline is 0.4. ( a ) Find the range of possible values for m 2...
View
Full
Document
This note was uploaded on 11/25/2010 for the course PHYSICS 4A taught by Professor Eduardo during the Spring '09 term at DeAnza College.
 Spring '09
 Eduardo
 Physics, Force

Click to edit the document details