HW2 Solutions

# HW2 Solutions - HOMEWORK SOLUTIONS Week 2 Due Friday 23...

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HOMEWORK SOLUTIONS Week 2: Due Friday 23 September 1. Find the reaction forces exerted on the bar AE at the pin A and at the support B in Figure 1 A B C 100 lb/ft 200 lb D E 8 5 2 2 all dimensions in feet. Figure 1 Solution Figure 1.1 shows a free-body diagram of the bar. We have shown a horizontal reaction R H at A , since the pin is capable of exerting such a force, but it is clear that it will actually be zero, since there are no other horizontal forces. A B C 100 lb/ft 200 lb D E 8 5 2 2 R B R A H R all dimensions in feet. Figure 1.1 Summing the moments about A , we have R B × 5 - 200 × 7 - 800 × 13 = 0 so R B = 2360 lb . Notice that in this equation, the resultant of the distributed force is 800 lb and it acts through the mid-point of the segment DE . 1

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Summing the forces vertically, we then have R A + R B - 200 - 800 = 0 so R A = 800 + 200 - 2360 = - 1360 lb . In other words, the reaction force at A is 1360 lb downwards . 2
2. Find the reaction forces exerted on the structure ABC by the pin at A and by the frictionless roller support at C . The members AB and AC are both of length 3 feet. A B C o 30 1500 lb Figure 2 Solution Figure 2.1 shows a free-body diagram of the structure. Notice that the pin at A is prevented from moving in any direction, so we put unknown horizontal and vertical reactions R 1 , R 2 . However, the pin at C is prevented from moving vertically, but not retrained horizontally, so we put just a vertical force R 3 . A B C o 30 1500 lb R 1 R 2 R 3 Figure 2.1 Summing the moments about A , we have R 3 × 3 - 1500 × 3 × sin30 = 0 or R 3 = 750 lb . 3

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Summing forces horizontally R 1 + 1500 = 0 so R 1 = - 1500 lb and summing forces vertically, R 2 + R 3 = 0 so R 2 = - R 3 = - 750 lb . Notice that we don’t try to guess the directions of the reaction forces. We put them all in the same direction (here in the positive x , y -directions) and let the algebra take care of the actual directions. 4
3. Find the largest force P that can be applied to the frame in Figure 3 if the resultant reaction at the pin B is not to exceed 5 kN. The cable AE passes around a frictionless pulley of radius 60 mm centered on C and is attached to the frame at E . A

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## This note was uploaded on 10/05/2011 for the course MECHENG 211 taught by Professor ? during the Fall '07 term at University of Michigan.

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HW2 Solutions - HOMEWORK SOLUTIONS Week 2 Due Friday 23...

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