Question

A uniform pole is propped between the floor and the ceiling of a room. The height of the room is 7.4 ft, and the coefficient of static friction between the pole an the ceiling is 0.565. The coefficient of static friction between the pole and the floor is greater than that between the pole and the ceiling.

For parts (a)-(e) assume the mass of the pole is 4.9 lbm and its length is 8.42 ft. Give your answers in pounds of force (lbf).

(a) Find the weight of the pole.
= _______  + ______ ẑ

(b) Find the normal force on the pole from the ceiling.
= _______  + ________ ẑ

(c) Find the frictional force from the ceiling.
= ________  + ________ ẑ

(d) Find the normal force from the floor
= _______  + ________ ẑ

(e) Find the frictional force from the floor
= ______  + ________ ẑ

Answer 1

(a) Given mass of the pole is 4.9lbm. The value of acceleration due to gravity g in FPS system is 32.2ft/s2.

Weight (W) of the pole is given by,

So the weight of the pole is

(b) The forces acting on the pole is given in the diagram below.

Given the height from floor to ceiling (h) = 7.4ft, the length of the pole (l) = 8.42ft and the coefficient of friction between ceiling and the pole () = 0.565. Let be the coefficient of friction between floor and the pole.

From the figure,

So the angle between floor and pole is 61.5 degree.

For the system to be in equilibrium the forces and the torque acting on the opposite directions should be equal. Taking forces in y-direction

Taking forces in x-direction,

Taking torque for the system (assuming clockwise torque positive and anti-clockwise torque negative) by taking floor as point of contact between floor and pole as point of rotation,

Dividing throughout by cos,

So the normal force on the pole from the ceiling is

(c) The frictional force from the ceilinng is f2 is given by

So the answer is

(d) The normal force from the floor (N1) is given by

So the answer is

(e) The frictional force from the floor (f1)

We know,

So the answer is