# phyexam1 - Version 109 – Midterm 1 – florin –(58140 1...

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Unformatted text preview: Version 109 – Midterm 1 – florin – (58140) 1 This print-out should have 18 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 10.0 points In the figure below the left-hand cable has a tension T 1 and makes an angle of 43 ◦ with the horizontal. The right-hand 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 , θ 1 = 43 ◦ , θ 3 = 46 ◦ , and T 2 = 58 N . T 3 T 1 θ 3 θ 1 T 3 cos θ 3 T 1 cos θ 1 W 2 W 1 Note: T 1 cos θ 1 = T 2 = T 3 cos θ 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 θ 3 acts up, so F net = W 2 − T 3 sin θ 3 = 0 = ⇒ T 3 sin θ 3 = W 2 . (1) Horizontally, T 2 acts to the left and T 3 cos θ 3 acts to the right, so F net = T 2 − T 3 cos θ 3 = 0 = ⇒ T 3 cos θ 3 = T 2 . (2) Dividing Eq. 1 by Eq. 2, we have tan θ 3 = W 2 T 2 . W 2 = T 2 tan θ 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 θ 1 − T 3 cos θ 3 = 0 , so W 1 = T 2 tan θ 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 μ =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 μ AB = 0 . 49 , between A and B μ 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 μ f μ The condition that block B not fall implies that its vertical acceleration is zero. Applying Newton’s second law for B in the horizontal and vertical directions yields summationdisplay F x = N = m B a x (1) summationdisplay F y = μ AB N − m B g = 0 , (2) where N is the normal force exerted on block B by cart A . Applying Newton’s law on cart A in the horizontal direction we have F −N = m A a x ....
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## This note was uploaded on 09/26/2010 for the course PHY 58230 taught by Professor Svhets during the Spring '10 term at University of Texas.

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phyexam1 - Version 109 – Midterm 1 – florin –(58140 1...

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