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 μs that will keep the pole horizontal and not allow its end to slide down the wall, we need to analyze the forces acting on the pole.
1. Let's start by considering the forces acting on the pole:
- The weight of the pole acts vertically downward at its center of mass, with a magnitude of Mg, where g is the acceleration due to gravity.
- The normal force on the pole acts vertically upward at the point of contact with the wall.
- The frictional force between the pole and the wall acts horizontally towards the left.
2. Since the pole is in equilibrium, the sum of the forces in the vertical direction must be zero:
Normal force - Mg = 0
=> Normal force = Mg
3. Now let's consider the torque (rotational force) acting on the pole:
The torque due to the weight of the pole is zero since it acts through the center of mass.
The torque due to the frictional force is also zero because the point of contact with the wall is the pivot point.
4. The torque due to the normal force can be calculated as follows:
Torque = normal force * distance from the pivot point
= Mg * (L/2)
= (MgL)/2
5. The torque due to the tension in the rope is:
Torque = tension * distance from the pivot point
= T * D
6. Since the pole is in equilibrium, the sum of the torques must also be zero:
(MgL)/2 + T * D = 0
7. Solve for T in terms of μs:
From points 3 and 5, we know T = μs * Normal force
Substituting T in the equation from point 6:
(MgL)/2 + μs * Normal force * D = 0
Substituting Normal force from point 2:
(MgL)/2 + μs * Mg * D = 0
8. Simplify the equation:
(MgL)/2 + μs * Mg * D = 0
(MgL)/2 = - μs * Mg * D
μs = - (MgL)/(2MgD)
9. Simplifying further:
μs = - L/(2D)
10. Final result:
The minimum value of μs is -L/(2D), where L is the length of the pole and D is the distance between the point of contact with the wall and the rope attachment point directly above it.
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 need to consider the equilibrium conditions.
Let's break down the forces acting on the pole:
1. Weight (mg): The entire mass of the pole creates a downward force (mg), acting at the center of mass. This force can be ignored from our analysis since it acts vertically and will not affect the minimum value of μs required.
2. Frictional force (Ff): The frictional force between the pole and the wall resists the downward sliding of the pole's end. This force acts in a vertical direction, opposing the component of the weight that wants to pull the pole down.
3. Tension force (T): The tension force in the rope supports the weight of the pole and provides an upward force at the other end of the pole.
Now, let's analyze the forces acting on the pole:
1. The vertical component of the weight (mg × sinθ) tries to rotate the pole clockwise.
2. The frictional force (Ff) tries to prevent the pole from sliding down the wall.
3. The tension force (T) tries to rotate the pole counterclockwise.
For the pole to remain in equilibrium and not rotate, the sum of the torques acting on it must be zero.
To simplify the analysis, let's consider the point where the pole contacts the wall as the pivot point. The torque due to the weight (mg × sinθ) can be represented as follows:
Torque due to weight = (mg × sinθ) × L/2
The torque due to the tension force (T) is represented as follows:
Torque due to tension = T × (L/2 + D)
Since the pole is in equilibrium, the sum of the torques must be zero:
(mg × sinθ) × L/2 - T × (L/2 + D) = 0
Rearranging this equation, we find:
μs × (M × g) × L/2 - T × (L/2 + D) = 0
Since we are interested in finding the minimum value of μs, we want to determine the situation where the static friction force (Ff) is at its minimum but still enough to prevent the pole from sliding down the wall. In this case, μs will be equal to its minimum value.
At the minimum value of static friction (Fmin), the frictional force (Ff) will be equal to the maximum force that static friction can provide, which is given by:
Fmax = μs × (M × g)
For the pole to remain in equilibrium, the tension force (T) must be equal to this maximum frictional force (Fmax):
T = Fmax
We can substitute Fmax and T into the equation above:
μs × (M × g) × L/2 - μs × (M × g) × (L/2 + D) = 0
Factoring out common terms and rearranging, we find:
μs × (M × g) × (L/2 - L/2 - D) = 0
μs × (M × g) × (-D) = 0
Since we want to determine the minimum value of μs, we assume D > 0. Therefore, the above equation reduces to:
-μs × (M × g) × D = 0
Finally, solving for μs, we have:
μs = 0
Thus, the minimum value of μs required to keep the pole horizontal and prevent its end from sliding down the wall is zero.