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

Question

A garage door is mounted on an overhead rail (the figure (Figure 1) ). The wheels at A and B have rusted so that they do not roll, but rather slide along the track. The coefficient of kinetic friction is 0.45. 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 weighs 994 N . It is pushed to the left at constant speed by a horizontal force F .

If the distance h is 1.71 m , what is the vertical component of the force exerted on the wheel A by the track?

If the distance h is 1.71 m , what is the vertical component of the force exerted on the wheel B by the track?

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 the constant speed". this means that the net force in the X-direction must be 0. so the force must equal the frictional force.

we have to find frictional force,

Frictional force = (normal force)*(Frictional force)

Normal force = weight of the door = 994N

F = 0.45 * 994 =447.3 N

a) the vertical component of the force exerted on the wheel A is

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*1.71 = W*1

Fa = 1/2(994-447.3 *1.71) = 114.5N

Since the net sum of the forces in the vertical must we know that Fb = 994 - Fa = 879.4 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.

b)Fb*2 = w*1 + F*1.71
Fb = (1/2)[994 + 447.3*1.71] = 879.4 N

c) maximum value h

Fa =1/2(W-F*h) =0

w =F*h

h =w/F =994/447.3 =2.2m.

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