**Unformatted text preview: **5.27: As in Example 5.17, the friction force is k n = k w cos and the component of the weight down the skids is w sin . In this case, the angle is arcsin(2.00 20.0) = 5.7. cos 25 The ratio of the forces is ksin = tank = 0..10 > 1, so the friction force holds the safe back, 0 and another force is needed to move the safe down the skids. b) The difference between the downward component of gravity and the kinetic friction force is w (sin - k cos ) = (260 kg) (9.80 m s 2 ) (sin 5.7 - (0.25) cos 5.7) = -381 N. ...

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