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HW3-phy2048-Spring-2014-solution

# 6 n 6p11 a the free body diagram for the crate

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Unformatted text preview: 0 m/s2 = 0.735m/s2 . a= m1 + m2 3.70 kg + 2.30 kg (b) The result for a is positive, indicating that the acceleration of block 1 is indeed up the incline and that the acceleration of block 2 is vertically down. (c) The tension in the cord is T = m1a + m1 g sin θ = (3.70 kg ) 0.735 m/s2 + (3.70 kg ) 9.80 m/s 2 sin 30.0° = 20.8N. ( ( ) ( ) ) 6P5. In addition to the forces already shown in Fig. 6- 17, a free- body diagram would r include an upward normal force FN exerted by the floor on the block, a downward ! mg representing the gravitational pull exerted by Earth, and an assumed- leftward ! f for the kinetic or static friction. We choose +x rightwards and +y upwards. We apply Newton’s second law to these axes: F − f = ma P + FN − mg = 0 where F = 6.0 N and m = 2.5 kg is the mass of the block. 4 PHY2048 Spring 2014 HW3 (a) In this case, P = 8.0 N leads to FN = (2.5 kg)(9.8 m/s2) – 8.0 N = 16.5 N. Using Eq. 6- 1, this implies f s ,max = µ s FN = 6.6 N , which is larger than the 6.0 N rightward force – so the block (which was initially at rest) does not move. Putting a = 0 into the first of our equations above yields a static friction force of f = P = 6.0 N. (b) In this case, P = 10 N, the normal force is FN = (2.5 kg)(9.8 m/s2) – 10 N = 14.5 N. Using Eq. 6- 1, this implies f s ,max = µ s FN = 5.8 N , which is less than the 6.0 N rightward force – so the block does move. Hence, we are dealing not with static but with kinetic friction, which Eq. 6- 2 reveals to be f k = µk FN = 3.6 N . 6P11. (a) The free- body diagram for the crate is shown on the ! r right. T is the tension force of the rope on the crate, FN ! is the normal force of the floor on the crate, mg is the ! force of gravity, and f is the force of friction. We take the +x direction to be horizontal to the right and the +y direction to be up. We assume the crate is motionless. The equations for the x and the y components of the force according to Newton’s second law are: T cos θ – f = 0 T sin θ + FN − mg = 0 where θ = 15° is the angle between the rope and the horizontal. The first equation gives f = T cos...
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