# midterm1 - Version 044 – Midterm 1 – florin –(58140 1...

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Unformatted text preview: Version 044 – 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 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 . . 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 = 94 N 2. F = 49 N 3. F = 64 N correct 4. F = 72 N 5. F = 86 N 6. F = 76 N 7. F = 82 N 8. F = 68 N 9. F = 84 N 10. F = 42 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 . Solving for F , F = N + m A a x . (3) The normal force N and the acceleration a x may be determined from the first two equa- tions. From Eq. 2 we find N = m B g μ AB . From Eq. 1 and 2, and solving for a x yields a x = N m B = g μ AB Since we now have expressions for N and a x , we can find the force F required so that block B does not fall. From Eq. 3 again F = N + m A a x = m B g μ AB + m A g μ AB = (2 . 3 kg + 0 . 9 kg) 9 . 8 m / s 2 . 49 = 64 N . Version 044 – Midterm 1 – florin – (58140) 2 002 10.0 points The suspended 2 . 7 kg mass on the right is moving up, the 1 . 4 kg mass slides down the ramp, and the suspended 8 . 2 kg mass on the left is moving down. There is friction between the block and the ramp. The acceleration of gravity is 9 . 8 m / s 2 . The pulleys are massless and frictionless. 1 . 4 k g μ = . 1 3 27 ◦ 8 . 2 kg 2 . 7 kg What is the tension in the cord connected to the 8 . 2 kg block? 1. 37.7017 2. 41.3336 3. 35.4779 4. 38.722 5. 33.3795 6. 38.1757 7. 44.2727 8. 36.3718 9. 43.6305 10. 39.6067 Correct answer: 41 . 3336 N. Explanation: Let : m 1 = 2 . 7 kg , m 2 = 1 . 4 kg , m 3 = 8 . 2 kg , and θ = 27 ◦ . Basic Concept: F net = ma negationslash = 0 Solution: The acceleration a of each mass is the same, but the tensions in the two strings will be different. Let T 1 be the tension in the right string and T 3 the tension in the left string. Consider the free body diagrams for each mass T 3 m 3 g a T 1 m 1 g a T 3 T 1 N μ N a m 2 g For the mass m 1 , T 1 acts up and the weight m 1 g acts down, with the acceleration a di- rected upward F net 1 = m 1 a = T 1 − m 1 g (1) For the mass on the table, the parallel compo- nent of its weight is...
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midterm1 - Version 044 – Midterm 1 – florin –(58140 1...

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