ch06-p013

# ch06-p013 - 13(a The free-body diagram for the crate is...

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Unformatted text preview: 13. (a) The free-body diagram for the crate is shown on the 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 s in θ + FN − mg = 0 where θ = 15° is the angle between the rope and the horizontal. The first equation gives f = T cos θ and the second gives FN = mg – T sin θ. If the crate is to remain at rest, f must be less than μs FN, or T cos θ < μs (mg – T sinθ). When the tension force is sufficient to just start the crate moving, we must have T cos θ = μs (mg – T sin θ). We solve for the tension: ( 0.50 ) ( 68 kg ) ( 9.8 m/s 2 ) μ s mg T= = = 304 N ≈ 3.0 ×102 N. cos θ + μ s sin θ cos 15° + 0.50 sin 15° (b) The second law equations for the moving crate are T cos θ – f = ma FN + T sin θ – mg = 0. Now f =μkFN, and the second equation gives FN = mg – Tsinθ, which yields f = μ k (mg − T sin θ ) . This expression is substituted for f in the first equation to obtain T cos θ – μk (mg – T sin θ) = ma, so the acceleration is a= Numerically, it is given by a= T ( cos θ + μk sin θ ) − μk g . m b304 Ngbcos15° + 0.35 sin 15°g − b0.35gc9.8 m / s h = 13 m / s . . 2 2 68 kg ...
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## This note was uploaded on 05/17/2011 for the course PHY 2049 taught by Professor Any during the Spring '08 term at University of Florida.

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