Ch06-p023 - θ − T 1 = 0 and the y component is T 2 sin θ − W A = 0 Eliminate the tension forces and find expressions for f and F N in terms

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23 The free-body diagrams for block B and for the knot just above block A are shown on the right. T 1 is the magnitude of the tension force of the rope pulling on block B , T 2 is the magnitude of the tension force of the other rope, f is the magnitude of the force of friction exerted by the horizontal surface on block B , F N is the magnitude of the normal force exerted by the surface on block B , W A is the weight of block A ,a n d W B is the weight of block B . θ (= 30 ) is the angle between the second rope and the horizontal. ..................................... . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .......... . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . T 1 f W B F N ................................................. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .......... . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . T 2 T 1 W A θ For each object take the x axis to be horizontal and the y axis to be vertical. The x component of Newton’s second law for block B is then T 1 f =0andthe y component is F N W B =0 . The x component of Newton’s second law for the knot is
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Unformatted text preview: θ − T 1 = 0 and the y component is T 2 sin θ − W A = 0. Eliminate the tension forces and find expressions for f and F N in terms of W A and W B , then select W A so f = µ s F N . The second Newton’s law equation gives F N = W B immediately. The third gives T 2 = T 1 / cos θ . Substitute this expression into the fourth equation to obtain T 1 = W A / tan θ . Substitute W A / tan θ for T 1 in the first equation to obtain f = W A / tan θ . For the blocks to remain stationary f must be less than µ s F N or W A / tan θ < µ s W B . The greatest that W A can be is the value for which W A / tan θ = µ s W B . Solve for W A : W A = µ s W B tan θ = (0 . 25)(711 N) tan 30 ◦ = 1 . × 10 2 N ....
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This note was uploaded on 05/13/2010 for the course PHYSICS 12121 taught by Professor Bill during the Winter '01 term at CSU Channel Islands.

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