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 855 N . It is pushed to the left at constant speed by a horizontal force F .

part a

If the distance h is 1.40 m , what is the vertical component of the force exerted on the wheel A by the track?

part b

If the distance h is 1.40 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

a)
Force F must equal the frictional force , because the door is moving at a constant speed, not accelerating.

Frictional force = (normal force)*(coefficient of friction) = F
Normal force = the weight of the door = 855 N
Coefficient of friction = 0.45
F = 0.45*855 = 384.75 N

Now 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. Fa and F are acting in the same direction (Clockwise), while Force due to Weight of door is acting opposite to it counter balancing it.
Calculating Forces /Torque around point B -
Fa * 2 + F *1.4 = w*1
Fa * 2 + 384.75 * 1.4 = 855
Fa = 158.175 N

b)
Now we know , Total force Downwards = Total force upwards as the system is stable and not accelearting
Therefore ,
Fa + Fb = Force due to Weight of door
Fb = 855 - 158.175 N
Fb = 696.825 N
c)
Put Fa= 0 in above equation of Torque around Point B .

Fa * 2 + F *h = w*1
0 * 2 + 384.75 * h = 855
h = 855/384.75
h = 2.22 m
maximum value h can have without causing one wheel to leave the track =2.22m

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