solution_pdf hw 4

# solution_pdf hw 4 - Patel(ppp285 – HW04 – TSOI...

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Unformatted text preview: Patel (ppp285) – HW04 – TSOI – (58160) 1 This print-out should have 20 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 (part 1 of 2) 10.0 points As shown in the figure, a block is pushed up against the wall. Let the mass of the block be m = 2 . 6 kg, the coefficient of kinetic friction between the block and the wall be μ = 0 . 63, and θ = 69 ◦ . Suppose F = 72 N. The acceleration of gravity is 9 . 8 m / s 2 . 2 . 6 kg F 6 9 ◦ μ k =0 . 63 Find the force of friction. Correct answer: 16 . 2555 N. Explanation: Recall that f = μN . From summationdisplay F = 0 in the horizontal direction, one can see that N = F cos θ . Hence the force of friction is f = μN = μF cos θ = 16 . 2555 N . 002 (part 2 of 2) 10.0 points The force, F , which keeps the block mov- ing upwards with a constant velocity satisfies which equations? 1. F cos θ = mg + μF sin θ 2. F cos θ = μmg 3. F sin θ = mg − μF sin θ 4. F cos θ = mg − μF cos θ 5. F cos θ = mg − μF sin θ 6. F sin θ = mg + μF sin θ 7. F cos θ = mg + μF cos θ 8. F sin θ = mg − μF cos θ 9. F sin θ = mg + μF cos θ correct 10. F sin θ = μmg Explanation: m F θ v mg f μ For constant velocity, acceleration is zero. Hence summationdisplay F y = 0 = F sin θ − mg − μF cos θ . The first term is the applied upward force, the second term is the weight of the block, and the third term is the frictional force. 003 10.0 points A 3.40 kg block is pushed along the ceiling with a constant applied force of 89.0 N that acts at an angle of 61.0 ◦ with the horizontal. The block accelerates to the right at 6.30 m/s 2 . The acceleration of gravity is 9 . 81 m / s 2 . 3 . 4 kg 8 9 N 6 1 ◦ μ 6 . 3 m / s 2 Patel (ppp285) – HW04 – TSOI – (58160) 2 What is the coefficient of kinetic friction between the block and the ceiling? Correct answer: 0 . 488412. Explanation: Basic Concepts: F applied,x = F applied cos θ F applied,y = F applied sin θ F y,net = F applied,y − mg − F n = 0 F x,net = ma x = F applied,x − F k F k = μ k F n m F θ μ a Given: m = 3 . 40 kg θ = 61 . ◦ F applied = 89 N a x = 6 . 30 m g = 9 . 81 m / s 2 Solution: F applied,x = (89 N) cos61 ◦ = 43 . 1481 N F applied,y = (89 N) sin61 ◦ = 77 . 8412 N The normal force is F n = F applied,y − mg = 77 . 8412 N − (3 . 4 kg)(9 . 81 m / s 2 ) = 44 . 4872 N From the horizontal motion, ma x = F applied,x − μ k F n μ k = F applied,x − ma x F n = 43 . 1481 N − (3 . 4 kg)(6 . 3 m / s 2 ) 44 . 4872 N = 0 . 488412 004 (part 1 of 3) 10.0 points The suspended 2 . 9 kg mass on the right is moving up, the 1 . 5 kg mass slides down the ramp, and the suspended 8 . 4 kg mass on the left is moving down. The coefficient of friction between the block and the ramp is 0 . 14 ....
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solution_pdf hw 4 - Patel(ppp285 – HW04 – TSOI...

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