ch06-p024 - 24. The free-body diagram for the block is...

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Unformatted text preview: 24. The free-body diagram for the block is shown below, with F being the force applied to the block, FN the normal force of the floor on the block, mg the force of gravity, and f the force of friction. We take the +x direction to be horizontal to the right and the +y direction to be up. The equations for the x and the y components of the force according to Newton’s second law are: Fx = F cos θ − f = ma Fy = FN − F sin θ − mg = 0 Now f =μkFN, and the second equation gives FN = mg + Fsinθ, which yields f = μk (mg + F sin θ ) . This expression is substituted for f in the first equation to obtain so the acceleration is F cos θ – μk (mg + F sin θ) = ma, a= F ( cos θ − μk sin θ ) − μk g . m From Fig. 6-32, we see that a = 3.0 m/s 2 when μk = 0 . This implies 3.0 m/s 2 = We also find a = 0 when μk = 0.20 : 0= F cos θ . m F F ( cos θ − (0.20) sin θ ) − (0.20)(9.8 m/s2 ) = 3.00 m/s2 − 0.20 sin θ − 1.96 m/s2 m m F 2 = 1.04 m/s − 0.20 sin θ m F sin θ . Combining the two results, we get m 5.2 m/s 2 tan θ = = 1.73 θ = 60°. 3.0 m/s 2 which yields 5.2 m/s 2 = ...
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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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