{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}


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

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: 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...
View Full Document

{[ snackBarMessage ]}

Ask a homework question - tutors are online