HW3 Solutions

# HW3 Solutions - HOMEWORK PROBLEMS Week 3 Due Friday 30...

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HOMEWORK PROBLEMS Week 3: Due Friday 30 September 1. In Figure 1, the disk is homogeneous and has a mass of 5 kg and the bar AB can be regarded as weightless. If the coefficient of friction at C is 0.2, what is the maximum downward force P that can be applied without slip. 200 200 120 140 A B C P 30 o all dimensions in mm Figure 1 Solution 200 200 120 140 A B P W N T 1 R 2 R 2 R 3 R 4 R Figure 1.1 Figure 1.1 shows a free-body diagram of the two components. Notice that for a given value of P , there are six unknowns ( R 1 ,R 2 ,R 3 ,R 4 ,N,T ) and six equations of equilibrium, three for each body. 1

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Taking moments about A for AB , we have R 2 × 400 - P × 200 = 0 so R 2 = P 2 . Taking moments about the center of the disk, we have R 2 × 120 - T × 140 = 0 so T = 120 R 2 140 = 6 P 14 . Summing the forces vertically for the disk, N cos(30) - T sin(30) - R 2 - W = 0 so N = 2 3 p 3 P 14 + P 2 + W P . For there to be no slip at C , we require T < μN or 6 P 14 < 0 . 4 3 p 3 P 14 + P 2 + W P or 6 P < 0 . 231(10 P + 14 W ) , giving P < 0 . 876 W = 0 . 876 × 5 × 9 . 81 = 42 . 97 N . 2
2. A small rectangular block rests without slipping on a plane inclined at 30 o to the horizontal. A horizontal force P is now applied to the block in the direction perpendicular to the line of greatest slope, as shown in Figure 2, and is gradually increased from zero. It is observed that the block starts to slide when P = 0 . 2 W where W is the weight of the block. What is the coefficient of friction between the block and the plane? P o 30 Figure 2 Solution Figure 2.1 shows a free-body diagram of the block looking in the direction of the applied force P . In this figure, the force F is that component of the friction force that acts in the direction up the inclined plane, but there must also be a component pointing out of the paper equal to the applied force P at the instant before slip occurs. W N F 30 o Figure 2.1 3

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Thus, the resultant frictional force is F 2 + P 2 . Summing the forces up the plane in Figure 2.1, we have F - W sin 30 = 0 so F = W 2 and the resultant frictional force is r p W 2 P 2 + (0 . 2 W ) 2 = 0 . 5385 W . Also, summing forces normal to the plane, N - W cos 30 = 0 so N = 0 . 866 W . We conclude that the coefficient of friction is μ = 0 . 5385 0 . 866 = 0 . 622 . 4
Figure 3 shows a manual automotive jack. Rotation of the screw at D causes C to be pulled towards D and hence AB to move upwards against the vehicle weight P = 3 kN. The dimensions AB and EF are both L = 100 mm and the diagonal members AD , BC , CF , DE are all of length 155 mm. Find the forces in all the two-force members, including that in the screw

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HW3 Solutions - HOMEWORK PROBLEMS Week 3 Due Friday 30...

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