A uniform rigid pole of length L and mass M is to be supported from a vertical wall in a horizontal position, as shown in the figure. The pole is not attached directly to the wall, so the coefficient of static friction, μs,between the wall and the pole provides the only vertical force on one end of the pole. The other end of the pole is supported by a light rope that is attached to the wall at a point a distance D directly above the point where the pole contacts the wall. Determine the minimum value of μs, as a function of L and D, that will keep the pole horizontal and not allow its end to slide down the wall. (Use any variable or symbol stated above as necessary.)
Assume the rope makes an angle theta with horizontal and tension in the rope is T. Let the normal reaction at the pole and wall contact point is N (in horizontal dir.)
T*Cos theta = N ....(1)
T*Sin theta + mu*N = Mg ........(2)
So, Tan theta = (M*g-mu*N)/N
or, Tan theta = M*g/N - mu
But Tan theta = D/L
So, M*g/N - mu = D/L
or, mu = M*g/N - D/L ......(3)
Now take moment of the forces about the point of contact:
T*(Sin theta)*L = M*g*L/2
or T*(Sin theta)= M*g/2 .....(4)
From (1)& (4) Tan theta = (M*g/2)/N
or, N = (M*g/2)/(D/L)
= M*g*L/2*D ........(5)
From (3) & (5)
mu = M*g*2D/M*g*L - D/L
= 2D/L - D/L
= D/L
Thank you so much! I've been trying to figure this one out for a few hours.
To determine the minimum value of μs, as a function of L and D, let's analyze the forces acting on the pole in equilibrium.
1. Consider the entire pole as a system, with a chosen pivot point. The net torque about this point must be zero for rotational equilibrium.
2. The weight of the pole acts at its center of mass, which is located at L/2 from either end. We can represent this force as a downward force of magnitude Mg acting vertically through the center of mass.
3. The force exerted by the rope supports the pole vertically. We can represent this force as an upward force of magnitude F.
4. The force of static friction provided by the wall acts perpendicular to it. We can represent this force as a horizontal force of magnitude f.
Now let's consider the torques. The torque due to the weight acts clockwise, whereas the torque due to the rope and friction acts counterclockwise.
- Torque due to weight (clockwise): Since the weight acts downward through the center of mass, its lever arm is L/2. The torque is given by (Mg)(L/2).
- Torque due to rope (counterclockwise): The lever arm is the distance D from the pivot point to the point where the rope is attached. The torque is given by (F)(D).
- Torque due to friction (counterclockwise): The lever arm is the length of the pole itself, L. The torque is given by (f)(L).
Since the pole is in equilibrium, the net torque must be zero:
(Mg)(L/2) + (F)(D) - (f)(L) = 0
To find the minimum value of μs, we need to consider the conditions for the pole just about to slide down the wall. At this point, the frictional force reaches its maximum value, which is given by:
f = μs*N
where N is the normal force acting on the pole, which is equal to Mg.
Substituting this expression for f in the torque equation and rearranging, we get:
(Mg)(L/2) + (F)(D) - (μs*Mg)(L) = 0
Now, we can solve this equation for μs:
(Mg)(L/2) - (μs*Mg)(L) = (F)(D)
Dividing through by Mg:
(L/2) - (μs)(L) = (F)(D)/(Mg)
Simplifying:
μs = (L/2D) + (F/Mg)
From this equation, we can see that the minimum value of μs depends on the ratio between L/2D and F/Mg. Specifically, as long as the sum of these ratios is less than or equal to 1, the pole will remain horizontal and not slide down the wall.