hw4a - HW#4a Page 1 of 4 Problem 1 = Walker P 5-23 To give...

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HW#4a Page 1 of 4 Problem 1 = Walker P. 5-23. To give a [17] kg child a ride, two teenagers pull on a 3.4 kg sled with ropes. Both teenagers pull with a force of F = 47 N at an angle of 35 ° relative to the forward direction, which is the direction of motion. In addition, the snow exerts a retarding force on the sled that points opposite to the direction of motion, and has a magnitude of 57 N. Find the size of the acceleration of the sled and child. Solution: There is lots of stuff shown on the picture. The two forward forces are symmetrical about the line of motion. So we can do this problem without much fuss. The net force forwards will be the sum of the forward components of the two ropes - each one is equal to Fx = F cos35 ° = (47 N) cos 35 ° = 38.50 N (each). The 57 N from friction acts backward. So the net forward force is forward 2*38.50 N - 57 N = 20 N F = and the acceleration is 2 2 20 kg-m/s 0.980 m/s 17 kg + 3.4 kg F a m == = If the above solution is a little risky to you, you should follow the problem solving strategy on lecture notes and do it step by step. Identify the object and there are 3 forces acting on it: F1, F2 and friction. Define the forward direction to be positive x, and define the north “or up” direction to be positive y. De-component the forces: F1 has horizontal component F1x = F1 cos35 ° y component: F1y= F1 sin 35 ° F2 has horizontal component F2x = F2 cos35 ° y component: F2y= - F2 sin 35 ° Because F1=F2 , these 2 forces have the same contribution in x direction and have the opposite contribution in y direction.
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This note was uploaded on 08/25/2010 for the course PHYS 111 taught by Professor Jamesm.lockhart during the Spring '09 term at S.F. State.

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hw4a - HW#4a Page 1 of 4 Problem 1 = Walker P 5-23 To give...

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