Question

Can someone clear up this misconception for me? I've posted part of the problem and my question below. The book is Vector Mechanics for Engineers Statics and Dynamics 8th edition

Problem 13.11 The 7-kg block A is released from rest in the position shown. Neglecting the effect of friction and the masses of the pulleys determine the velocity of the block after it has moved 0.6 m up the incline 1.2 I 30 15° 140 N Fig, P13.11 Given: Block A is released from rest and moves up incline 0.6 m. Friction and other masses are neglected Find: Velocity of the block after 0.6 m, v From the Law of Cosines d2- (1.2)2+ (0.6)2 - 2(1.2) (0.6)cos 15° d2- 0.4091 m2 d 0.63958 m 15 Uc-Wc (Distance pulley C lowered) WA 7(9.81) 68.67 N 140 N1.2 0.63958)m -39.229 N-m Question: In solving this problem 0.6m is used for the length of one segment and 1.2m is used for another. With this information, we can use law of cosines to find the length the side opposite the 15° angle segment BC. What I don't understand is how 0.6 and 1.2 can be used to solve for the unknown side as they, to my understanding, exist in two different states. For example, segment AB is 1.2m long before block A travels 0.6m, so how can 1.2 be used in the law of cosines equation along with 0.6m?

Answer 1

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