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.)
To determine the minimum value of the coefficient of static friction, μs, that will keep the pole horizontal and prevent its end from sliding down the wall, we can consider the equilibrium conditions for the pole.
Let's analyze the forces acting on the pole:
1. The weight of the pole acts downward at its center of mass, which is located at L/2 from the supported end. This weight can be represented by the force Mg, where g is the acceleration due to gravity.
2. The reaction force at the supporting end of the pole, which can be resolved into its vertical and horizontal components.
3. The vertical component of the reaction force balances the weight of the pole:
Rv = Mg
4. The horizontal component of the reaction force, in conjunction with the frictional force, counteracts the tendency for the pole to rotate:
Rh + μsRv = 0
5. The torque about the point where the pole contacts the wall should be zero for rotational equilibrium:
(D-L/2)Rh - (L/2)Rv = 0
Now, let's solve these equations to find the minimum value of μs:
From equation (3), we have Rv = Mg
Substituting this into equation (4), we get:
Rh + μs(Mg) = 0
Simplifying, we have:
Rh = -μs(Mg)
Substituting Rh = -μs(Mg) into equation (5), we get:
(D - L/2)(-μs)(Mg) - (L/2)(Mg) = 0
Expanding and rearranging, we have:
DμsMg - (L/2)μsMg - (L/2)Mg = 0
Factoring out Mg and dividing by Mg, we get:
Dμs - (L/2)μs - (L/2) = 0
Now, let's isolate μs:
μs(D - L/2) = (L/2)
μs = (L/2)/(D - L/2)
Therefore, the minimum value of μs, as a function of L and D, is given by:
μs = (L/2)/(D - L/2)
To determine the minimum value of the coefficient of static friction (μs) needed to keep the pole in a horizontal position without sliding down the wall, we can start by analyzing the forces acting on the pole.
In this scenario, the pole is acted upon by its weight (mg) acting downwards at its center of mass. The horizontal force in equilibrium is achieved by the frictional force (Fs) between the pole and the wall, acting in response to the force applied by the rope.
Let's break down the forces acting on the pole:
1. Weight (mg): This force acts vertically downwards at the center of mass of the pole.
2. Frictional force (Fs): This force acts horizontally to balance the vertical force applied by the rope. Fs is equal and opposite to the vertical force Fv exerted by the rope.
3. Tension force (T): The rope pulling upwards on the pole at an angle. This force has both horizontal and vertical components.
- The horizontal component (Tcosθ) does not affect the equilibrium of the pole, as it cancels out with the frictional force.
- The vertical component (Tsinθ) provides the necessary force to counterbalance the weight of the pole.
Now, let's consider the torques acting on the pole. Since the pole is in a horizontal position, the torques must balance each other around the point of contact with the wall.
The torque due to weight (mg) produces a clockwise rotation, while the torque due to the vertical component of force (Tsinθ) produces a counterclockwise rotation.
For a system to remain in equilibrium, the sum of the torques must be equal to zero.
The torque due to weight (mg) can be calculated as the product of the weight force and the distance from the point of contact with the wall to the center of mass of the pole (L/2).
The torque due to the vertical component of force (Tsinθ) can be calculated as the product of the vertical component of tension and the distance from the point of contact with the wall to the point where the rope is attached (D).
Setting up the equation for torque equilibrium:
(mg)(L/2) = (Tsinθ)(D)
Simplifying the equation:
mgL = 2TsinθD
Now, since the rope is assumed to be light, we can ignore its mass and consider it as a tension force acting at the end of the pole (point A). The tension force can be expressed as T = μs(N), where N is the normal force exerted by the pole on the wall. Considering the pole as a beam, N is equal to mg (the weight of the pole) acting at its center of mass.
Substituting T = μs(N) into the torque equation:
mgL = 2(μs(N))(sinθ)D
Simplifying further:
mgL = 2μs(mg)(sinθ)D
The mass of the pole (m) cancels out:
L = 2μs(sinθ)D
Now, we need to determine the relationship between sinθ and the lengths L and D. To do this, let's consider the right triangle formed by the pole, the vertical line from the point of contact to the top of the pole, and the hypotenuse (the rope).
In this triangle, sinθ = L/D.
Replacing sinθ with L/D in the equation:
L = 2μs(L/D)D
Simplifying:
L = 2μsL
Now, we can rearrange the equation to solve for the minimum value of μs:
1 = 2μs
Dividing both sides by 2:
μs = 1/2
Therefore, the minimum value of the coefficient of static friction (μs) required to keep the pole horizontal without sliding down the wall is 1/2.