The problem can be simplified as a box slide down a hill as shown in the figure on the left. The center of gravity is 3 feet from the bottom.
There are three possible forces acting on the box, its weight, reaction force due to the ground and the friction force. The location and direction of these forces are shown on the free-body diagram of the box on the left. Let the resultant normal force, N be acting at point A. There are two unknowns, N and Fs.
Note that the co-ordinate system is rotated by 22.5o in clockwise direction to simplify the calculations. For equilibrium,
ΣFy = 0
N - w cos 22.5 = 0
N = w cos 22.5
ΣFx = 0
w sin 22.5 - Fs = 0
Fs = μ N = μ w cos 22.5
Putting back Fs in ΣFx equation gives,
w sin 22.5 = μ w cos 22.5
μ = tan 22.5 = 0.4142
Notice from the calculation that static coefficient of friction, μ, is independent of the weight of a box. Also, knowing the inclination angle of the bed is a convenient way of measuring the coefficient of static friction. |