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Unformatted text preview: This printout should have 23 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 40 ◦ with the horizontal. The righthand cable has a tension T 3 and makes an angle of 49 ◦ 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 77 N and is horizontal. The acceleration of gravity is 9 . 8 m / s 2 . M 1 M 2 77 N T 1 T 3 4 9 ◦ 4 ◦ Determine the mass M 2 . Correct answer: 9 . 03861 kg. Explanation: Given : W 1 = M 1 g , W 2 = M 2 g , θ 1 = 40 ◦ , θ 3 = 49 ◦ , and T 2 = 77 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 = (77 N) tan49 ◦ = 117 . 367 N M 2 = W 2 g = 88 . 5784 N 9 . 8 m / s 2 = 9 . 03861 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 = (77 N) tan40 ◦ = 100 . 516 N M 1 = W 1 g = 64 . 6107 N 9 . 8 m / s 2 = 6 . 59293 kg . 002 10.0 points A block of mass 2 . 59 kg lies on a frictionless horizontal surface. The block is connected by a cord passing over a pulley to another block of mass 7 . 97 kg which hangs in the air, as shown on the following picture. Assume the cord to be light (massless and weightless) and unstretchable and the the pulley to have no friction and no rotational inertia. The acceleration of gravity is 9 . 8 m / s 2 . 2 . 59 kg 7 . 97 kg oldmidterm2 02 – Turner – (58185) 1 hinojosa (jlh3938) – oldmidterm2 02 – Turner – (58185) 2 Calculate the acceleration of the first block. Correct answer: 7 . 3964 m / s 2 . Explanation: Let : m 1 = 2 . 59 kg and m 2 = 7 . 97 kg . m 1 m 2 a T N m 1 g a T m 2 g Since the cord is unstretchable, the first block accelerates to the right at exactly the same rate a as the second (hanging) block ac celerates downward. Also, the cord’s tension pulls the first block to the right with exactly the same tension T as it pulls the second block upward. The only horizontal force acting on the first block is the cord’s tension T , hence by New ton’s Second Law m 1 a = F net → 1 = T . The second block feels two vertical forces: The cord’s tension T (upward) and the block’s own weight W 2 = m 2 g (downward). Conse quently, m 2 a = F net ↓ 2 = m 2 g T ....
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 Spring '08
 Turner
 Force, Friction, Correct Answer, kg, Hinojosa

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