Physics Solution Manual for 1100 and 2101

This force is part of block 1 the action reaction

Info iconThis preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: uation (1) for the airplane’s acceleration: ax = µs mg cos θ T cos θ = M M ( cos θ − µs sin θ ) (6) SOLUTION We apply Equation (6) to calculate the acceleration of the airplane: ( ) ( 0.77 )( 85 kg ) 9.80 m/s cos 9.0 µs mg cos θ = = 6.7 × 10−3 m/s 2 ax = o o M ( cos θ − µs sin θ ) (109 000 kg ) cos 9.0 − 0.77 sin 9.0 ( 2 o ) Chapter 4 Problems 87. REASONING As the free-body diagram shows, there are two forces acting on the fireman as he slides down the pole: his weight W and the kinetic frictional force fk. The kinetic frictional force opposes the motion of the fireman, so it points upward in the +y direction. In accord with Newton’s second law, the net force, which is the sum of these two forces, is equal to the fireman’s mass times his acceleration. His mass, and therefore his weight, is known, but his acceleration is not. We will turn to one of the equations of kinematics from Chapter 3 to determine the acceleration. SOLUTION Newton’s second law ( ΣFy = ma y , Equation 4.2b ) can 219 +y fk W Free-body diagram for the fireman be applied to this situation: ΣFy = + f k − W = ma y The magnitude W of the fireman’s weight can be expressed in terms of his mass as W = mg (Equation 4.5), where g is magnitude of the acceleration due to gravity. Solving the equation above for the magnitude of the kinetic frictional force, and using W = mg, gives f k = ma y + W = ma y + mg (1) Since the initial and final velocities, v0y and vy, and the displacement y are known, we will use Equation 3.6b from the equations of kinematics to relate these variables to the 2 acceleration: v 2 = v0 y + 2a y y . Solving this equation for ay and substituting the result into y Equation 1 gives 2 v 2 − v0 y y fk = m 2y + mg We note that the fireman slides down the pole, so his displacement is negative, or y = −4.0 m. The magnitude of the kinetic frictional force is, then, 2 v 2 − v0 y y fk = m 2y + mg (1.4 m/s )2 − ( 0 m/s )2 = ( 86 kg ) + ( 86 kg ) ( 9.80 m/s 2 ) = 820 N 2 ( −4.0 m ) ______________________________________________________________________________ 220 FORCES AND NEWTON'S LAWS OF MOTION 88. REASONING Consider the forces that act on each block. Only one + − force contributes to the horizontal net force acting on block 1, as shown in the free-body diagram. This is the −P force −P with which block 2 pushes on block 1. The minus sign in the free-body diagram indicates the Free-body diagram for direction of the force is to the left. This force is part of block 1 the action-reaction pair of forces that is consistent with Newton’s third law. Block 1 pushes forward and to the right against block 2, and block 2 pushes backward and to the left against block 1 with an oppositely directed force of equal magnitude. Two forces contribute to the horizontal net force acting + − on block 2, as shown in the free-body diagram. One is the force P with which block 1 pushes on block 2. P fk According to Newton’s third law, this force has the same magnitude but the opposite direction as the force Free-body diagram for with which block 2 pushes on block 1. The other force block 2 is the kinetic frictional force fk, which points to the left, in opposition to the relative motion between the block and the surface on which it slides. Both blocks decelerate, the magnitude of the deceleration being the same for each block. They have the same deceleration, because they are pressed together. Since the blocks are moving to the right in the drawing, the acceleration vector points to the left, for it reflects the slowing down of the motion. In Case A and in Case B we will apply Newton’s second law separately to each block in order to relate the net force to the acceleration. SOLUTION Referring to the free-body diagram for block 1, we write Newton’s second law as follows: −P (1) { = m1 ( −a ) Net force on block 1 where a is the magnitude of the acceleration. The minus sign appears on the right side of this equation because the acceleration, being a deceleration, points to the left, in the negative direction. Referring to the free-body diagram for block 2, we write Newton’s second law as follows: (2) P − f k = m2 ( − a ) 13 2 Net force on block 2 Solving Equation (1) for a gives a = P / m1 . Substituting this result into Equation (2) gives −P P − f k = m2 m 1 or P= m1 f k m1 + m2 Chapter 4 Problems 221 Substituting this result for P into a = P / m1 gives a= m1 f k fk P = = m1 m1 ( m1 + m2 ) m1 + m2 We can now use these results to calculate P and a in both cases. a. Case A Case B m1 f k P= b. Case A = ( 3.0 kg ) ( 5.8 N ) = 2.9 N m1 f k P= = ( 6.0 kg )( 5.8 N ) = 3.9 N m1 + m2 m1 + m2 − fk a= m1 + m2 = 3.0 kg + 3.0 kg 6.0 kg + 3.0 kg −5.8 N = −0.97 m/s 2 3.0 kg + 3.0 kg The magnitude of the acceleration is 0.97 m/s 2 . Case B a= − fk m1 + m2 = −5.8 N = −0.64 m/s 2 6.0 kg + 3.0 kg The magnitude of the acceleration is 0.64 m/s 2 . 89. SSM REASONING The tension in each coupling bar is responsible for accelerating the objects behind it. The masses of the cars are m1, m2, and m3....
View Full Document

Ask a homework question - tutors are online