ch06-p026 - 26. The free-body diagram for the sled is shown...

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Unformatted text preview: 26. The free-body diagram for the sled is shown on the right, with F being the force applied to the sled, FN the normal force of the inclined plane on the sled, mg the force of gravity, and f the force of friction. We take the +x direction to be along the inclined plane and the +y direction to be in its normal direction. The equations for the x and the y components of the force according to Newton’s second law are: Fx = F − f − mg sin θ = ma = 0 Fy = FN − mg cos θ = 0 Now f =μFN, and the second equation gives FN = mgcosθ, which yields f = μ mg c os θ . This expression is substituted for f in the first equation to obtain F = mg (sin θ + μ cos θ ) From Fig. 6-34, we see that F = 2.0 N when μ = 0 . This implies mg sin θ = 2.0 N. Similarly, we also find F = 5.0 N when μ = 0.5 : 5.0 N = mg (sin θ + 0.50 cos θ ) = 2.0 N + 0.50mg cos θ which yields mg cos θ = 6.0 N. Combining the two results, we get tan θ = 21 = 63 θ = 18°. ...
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