A garage door is mounted on an overhead rail (the figure (Figure 1)). The wheels

Question

A garage door is mounted on an overhead rail (the figure (Figure 1)). The wheels at A and B have rusted so that do not roll, but rather slide along the track. The coefficient of kinetic friction is 0.47. The distance between the wheels is 2.00 m, and each is 0.50 m from the vertical sides of the door. The door is uniform and weight 987 N. It is pushed to the left at constant speed by a horizontal force F. If the distance h is 1.68 m , what is the vertical component of the force exerted on the wheel A by the track? Part B If the distance h is 1.68 m , what is the vertical component of the force exerted on the wheel B by the track? Part C Find the maximum value h can have without causing one wheel to leave the track.

Explanation / Answer

  Note that the door is "pushed to the left at constant speed". This means that the net force in the X direction must be 0 (otherwise the door would be accelerating). So the force F must equal the frictional force.

Frictional force = (normal force)*(coefficient of friction) = F
Normal force = the weight of the door = 987 N
Coefficient of friction = 0.47
F = 0.47*987 = 463.89 N

Since the door is not accelerating in the vertical, there must be forces acting upwards on A and B to balance the force of gravity pulling down. The forces act on A and B since they are the points of contact with the rail the door slides on.

Look at the point B and the torque on it. The force at B is not important since it is acting at the point and therefore produces no torque. This leaves the weight of the door (which acts at the center of mass of the door), the force F and the force Fa at A. The net torque must be zero since the door is not rotating.

Fa and F are acting in the same direction, clockwise as the diagram as drawn, and the sum of their torques must equal the torque from the weight of the door which is acting in a counterclockwise direction.
Fa*2 + F*1.68 = w*1
Fa = (1/2)[w - F*1.68] = (1/2)[987 - 463.89*1.68] = 103.8324 N

Since the net sum of the forces in the vertical must we know that Fb = 987 - Fa = 883.1676 N but we can go ahead and do a similar analysis as we did for the point B. At A, F and the weight of the door are acting in a clockwise direction and Fb is acting counterclockwise.

Fb*2 = w*1 + F*1.68
Fb = (1/2)[987 + 463.89*1.68] = 883.1676 N

We now want to know at what distance h one of the wheels will leave the track. This will be B since a larger vertical force is acting on it. You notice how the weight of the door is distributed between A and B and the maximum that either of these can be will be equal to the total weight of the door. So B will reach this point when the force at A is 0 and there is no longer any contribution to the torque from this force. So in the equation for troque around B we need to set Fa to zero and then solve for the distance to F as a variable.

Fa = (1/2)[w - F*h] = 0
w = F*h and h = w/F = 987/463.89 = 2.12766 m

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