A 10-lb collar is attached to a spring and slides without friction along a fixed

Question

A 10-lb collar is attached to a spring and slides without friction along a fixed rod in a vertical plane. The spring has an undeformed length of 14 in. and a constant k = 4 lb/in. Knowing that the collar is released from rest in the position shown, determine the force exerted by the rod on the collar at

(a) Point A

(b) Point B

Both of these points are on the curved portion of the rod.

I found the length of the spring at the starting position, Point A, and Point B. I am having difficulty figuring out what else to do. Please help!

Explanation / Answer

You have find the length of the spring at point A which comes out to be 19.79 in. in this case,then find the elongation at point A, then we get the force exerted by the spring on the collar = F_s = k*(deltax_A) [ 0.707 (i) + 0.707 ( j) ] = 4*5.79 [ 0.707 (i) + 0.707 ( j) ] Force of gravity is acting downwards F_g = m_c*g ( - j) the force exerted by the rod on the collar at pt. A should be opposite to the sum of the F_s and F_g in the vertical direction so that the collar remains in equilibrium in the vertical direction. therefore, the force exerted by the rod on the collar at pt. A = - (16.37 - 322) (j) = 305.63 in the downward direction At point B ,there will be no force by the rod on the collar because there is no movement of collar in the direction perpendicular to the rod at point B

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