A worker drags a crate across a factory floor by pulling on a rope tied to the c

Question

A worker drags a crate across a factory floor by pulling on a rope tied to the crate (Fig. 6-60). The worker exerts a force of 450 N on the rope, which is inclined at 38degree to the horizontal. and the floor exerts a horizontal force of 125 N that opposes the motion. Calculate the magnitude of the acceleration of the crate if (a) its mass is 310 kg or (b) its weight is 310 N.

Explanation / Answer

First you should draw a free-body diagram to determine the forces acting on the crate. In this problem, there are only two forces acting horizontally on the crate: the horizontal component of the external force and the frictional force. Use Newton's 2nd law to determine an equation to solve for the acceleration: External force (horizontal component) - Frictional Force = m*a Since the external force is at an angle, we only need the horizontal component because that's the part of the force that moves the crate. Use trigonometry to find this value. External force(horizontal) = F*cos(20) = 360*cos(20) = 338.3 N Now solve for the frictional force: Frictional force = u*N, since N = weight of block, this simplies to: Fricitional force = (0.25)(1000) = 250 N Now, we can solve for the net force: External force (horizontal component) - Frictional Force = m*a 338.3 - 250 = m*a, thus: 88.3 = m*a To solve for the mass, use the given weight: W = m*g ----------> m = W/G = 1000/[9.8 m/s^2] = 102.04 kg Now we can solve for the acceleration: 88.3 N = (102.04 kg)*a --------> a = 0.87 m/s^2 Hope this helps

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