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HW 005 - hanna(brh687 – hk 5 – Opyrchal –(11105 1...

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Unformatted text preview: hanna (brh687) – hk 5 – Opyrchal – (11105) 1 This print-out should have 15 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 The system is balanced so that the 55 N block on the left and the 10 kg block on the right don’t quite slip. The suspended block has a mass of 3 kg and the angle is 41 ◦ . The acceleration of gravity is 9 . 8 m / s 2 . 55 N μ 1 10 kg μ 2 3 kg 4 1 ◦ Find the coefficient of friction μ 1 for the 3 kg block for this to be true. Correct answer: 0 . 614924. Explanation: Let : θ = 41 ◦ , m 1 = 3 kg , m 2 = 10 kg , W = 55 N , and g = 9 . 8 m / s 2 . Basic Concepts At equilibrium, summationdisplay F net = ma = 0 both vertically and horizontally. Solution Let T 1 be the tension in the left string and T 2 be the tension in the right string. Consider the free body diagrams for W , the point of attachment of the three strings, and the mass m 2 T 1 μ 1 N 1 N 1 W 55 N T 2 T x 2 T y 2 μ 1 N 1 m 1 g μ 2 N 2 T 2 N 2 m 2 g 10 kg For the weight W , the weight W acts down, the normal force N 1 acts up, the tension T 1 acts to the right and friction μ 1 N 1 acts to the left so that vertically F net = N 1-W = 0 N 1 = W and horizontally F net = T 1- μ 1 N 1 = 0 T 1 = μ 1 W . (1) For the point of attachment of the three strings, the weight m 1 g acts down, T 1 acts to the left, T 2 sin θ acts up and T 2 cos θ acts to the right so that vertically F net = m 1 g- T 2 sin θ = 0 T 2 = m 1 g sin θ , (2) and horizontally F net = T 1- T 2 cos θ = 0 T 1 = T 2 cos θ . (3) For the mass m 2 , the weight m 2 g acts down, N 2 acts up, T 2 acts to the left and friction μ 2 N 2 acts to the right so that vertically F net = m 2 g-N 2 = 0 N 2 = m 2 g , hanna (brh687) – hk 5 – Opyrchal – (11105) 2 and horizontally F net = T 2- μ 2 N 2 = 0 T 2 = μ 2 m 2 g . (4) From Eqs. 1, 2, and 3, we obtain μ 1 = T 1 W = T 2 cos θ W = m 1 g cos θ W sin θ = (3 kg) ( 9 . 8 m / s 2 ) cos 41 ◦ (55 N) sin 41 ◦ = . 614924 . 002 (part 2 of 2) 10.0 points Find the coefficient of friction μ 2 for the 10 kg block for this to be true. Correct answer: 0 . 457276. Explanation: From Eqs. 2 and 4, we have μ 2 = T 2 m 2 g = m 1 g m 2 g sin θ = m 1 m 2 sin θ = (3 kg) (10 kg) sin 41 ◦ = . 457276 . 003 10.0 points At the beginning of a new school term, a student moves a box of books by attaching a rope to the box and pulling with a force of F = 95 . 7 N at an angle of 56 ◦ , as shown in the figure. The acceleration of gravity is 9 . 8 m / s 2 . The box of books has a mass of 15 kg and the coefficient of kinetic friction between the bottom of the box and the floor is 0 . 49....
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HW 005 - hanna(brh687 – hk 5 – Opyrchal –(11105 1...

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