A toy chest and its contents have a combined weight of W = 180 N. The coefficien

Question

A toy chest and its contents have a combined weight of W = 180 N. The coefficient of static friction between toy chest and floor mu_s is 0.460. The child in the figure attempts to move the chest across the floor by pulling on an attached rope. (a) If theta is 45.0 degree, what is the magnitude of the force F that the child must exert on the rope to put the chest on the verge of moving? Determine (b) the value of theta for which F is a minimum and (c) that minimum magnitude. (a) Number units (b) Number units (c) Number units

Explanation / Answer

W=180N, = 0.46, theta = 45,

(a) Vertical component of force = F sin theta = 0.7F

Horizontal component of force = F cos theta = 0.7F

Assuming all forces to be acting at the centr of mass, 180N-0.7F will be the contact force and (180N-0.7F) is the frictional force. To bring the chest to the verge of moving, horizontal moving force and frictional force are equal, or, (180N-0.7F) = 0.7F

or, 0.46(180-0.7F)=0.7F

or, F=80.25N

(b) Now, as the minimum force is required only when the angle is eual to , which is the angle of friction and obatined by tan-1 =  tan-10.46 = 24.7.

The angle theta is 24.7 for minimum force.

(c) When theta = 24.7, (W - F sin theta) = F cos theta

or, 0.46(180-0.417F)=0.908F

or, F = 75.2N

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