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Unformatted text preview: Version 109 Midterm 1 florin (58140) 1 This printout should have 18 questions. Multiplechoice questions may continue on the next column or page find all choices before answering. 001 10.0 points In the figure below the lefthand cable has a tension T 1 and makes an angle of 43 with the horizontal. The righthand cable has a tension T 3 and makes an angle of 46 with the horizontal. A W 1 weight is on the left and a W 2 weight is on the right. The cable connecting the two weights has a tension 58 N and is horizontal. The acceleration of gravity is 9 . 8 m / s 2 . M 1 M 2 58 N T 1 T 3 4 6 4 3 Determine the mass M 2 . 1. 8.77 2. 7.82 3. 9.58 4. 7.36 5. 7.16 6. 7.61 7. 6.13 8. 4.54 9. 3.6 10. 4.75 Correct answer: 6 . 13 kg. Explanation: Given : W 1 = M 1 g , W 2 = M 2 g , 1 = 43 , 3 = 46 , and T 2 = 58 N . T 3 T 1 3 1 T 3 cos 3 T 1 cos 1 W 2 W 1 Note: T 1 cos 1 = T 2 = T 3 cos 3 Basic Concepts: summationdisplay vector F = mvectora = 0 vector W = mvectorg Solution: Consider the point of attachment of cable 2 and cable 3. Vertically, W 2 = M 2 g acts down and T 3 sin 3 acts up, so F net = W 2 T 3 sin 3 = 0 = T 3 sin 3 = W 2 . (1) Horizontally, T 2 acts to the left and T 3 cos 3 acts to the right, so F net = T 2 T 3 cos 3 = 0 = T 3 cos 3 = T 2 . (2) Dividing Eq. 1 by Eq. 2, we have tan 3 = W 2 T 2 . W 2 = T 2 tan 3 = (58 N) tan46 = 60 . 0608 N M 2 = W 2 g = 60 . 0608 N 9 . 8 m / s 2 = 6 . 13 kg and by symmetry , we have T 1 cos 1 T 3 cos 3 = 0 , so W 1 = T 2 tan 1 = (58 N) tan43 = 79 . 305 N M 1 = W 1 g = 54 . 0859 N 9 . 8 m / s 2 = 5 . 52 kg . 002 10.0 points Static friction 0 . 49 between a 0 . 9 kg block and a 2 . 3 kg cart. There is no kinetic friction between the cart and the horizontal surface. The acceleration of gravity is 9 . 8 m / s 2 . Version 109 Midterm 1 florin (58140) 2 . 9 kg F 2 . 3 kg =0 . 49 9 . 8m / s 2 What minimum force F must be exerted on the 2 . 3 kg cart in order for the 0 . 9 kg block not to fall? 1. F = 92 N 2. F = 88 N 3. F = 74 N 4. F = 68 N 5. F = 98 N 6. F = 66 N 7. F = 96 N 8. F = 48 N 9. F = 64 N correct 10. F = 84 N Explanation: Let : m A = 2 . 3 kg , Cart A m B = 0 . 9 kg , Block B AB = 0 . 49 , between A and B k = 0 , horizontal surface , and g = 9 . 8 m / s 2 . m A = 2 . 3 kg m B = . 9 kg F N BA N AB m A g m B g f f The condition that block B not fall implies that its vertical acceleration is zero. Applying Newtons second law for B in the horizontal and vertical directions yields summationdisplay F x = N = m B a x (1) summationdisplay F y = AB N m B g = 0 , (2) where N is the normal force exerted on block B by cart A . Applying Newtons law on cart A in the horizontal direction we have F N = m A a x ....
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 Spring '10
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