ch12-p039 - 39. The force diagram shown below depicts the...

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11 1.2m tan tan 33.7 . 2 2(0.90m) h θ −− == = ° A As is increased from zero the crate slides before it tips. (b) It starts to slide when = 31º. (c) The crate begins to slide when = tan –1 μ s = tan –1 0.70 = 35.0º and begins to tip when = 33.7º. Thus, it tips first as the angle is increased. (d) Tipping begins at = 33.7 °≈ 34 ° . 39. The force diagram shown below depicts the situation just before the crate tips, when the normal force acts at the front edge. However, it may also be used to calculate the angle for which the crate begins to slide. W is the force of gravity on the crate, N F is the normal force of the plane on the crate, and f is the force of friction. We take the x axis to be down the plane and the y axis to be in the direction of the normal force. We assume the acceleration is zero but the crate is on the verge of sliding. (a) The x and y components of Newton’s second law are sin 0 and cos 0 N Wf F W θθ −= = respectively. The y equation gives F N = W cos . Since the crate is about to slide f = s F N = s W cos , where s is the coefficient of static friction. We substitute into the
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