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.)
In order to determine the minimum value of the coefficient of static friction μs, we need to analyze the forces acting on the pole.
Let's consider the forces acting on the pole:
1. The weight of the pole, which acts vertically downward at its center of mass. The magnitude of this force is given by Fg = Mg, where M is the mass of the pole and g is the acceleration due to gravity.
2. The normal force exerted by the wall on the pole, which acts perpendicular to the wall and supports the weight of the pole. This force is denoted by FN.
3. The force of static friction between the pole and the wall, which acts vertically upward at the end of the pole. This force is denoted by Ff.
In order for the pole to be in equilibrium and remain horizontal, the sum of the moments about any point must be zero.
Let's take moments about the point where the pole touches the wall. The moment due to the weight of the pole about this point is zero since it acts along the line of action passing through the point.
The moment due to the normal force FN is also zero, as it acts vertically upwards through the point of contact.
The moment due to the force of static friction Ff is non-zero and tends to rotate the pole clockwise. The magnitude of this moment is given by MFf = Ff * L/2, where L/2 is the distance from the point of contact to the center of mass of the pole.
For the pole to remain in equilibrium, the moment due to the force of static friction must be balanced by the moment due to the weight of the pole. Therefore, we have:
MFf = Mg * (L/2)
Substituting the expression for MFf and rearranging, we get:
Ff * L/2 = Mg * (L/2)
Ff = Mg
Since Ff represents the force of static friction, we can replace it with μs * FN, where FN is the normal force.
μs * FN = Mg
To solve for the minimum value of μs, we need to determine the minimum value of FN.
From the given figure, we can observe that the tension in the rope creates a torque about the point of contact. The magnitude of this torque is given by TD = (D/L) * Mg, where D/L is the distance ratio.
In order to counterbalance this torque and keep the pole horizontal, the static friction force must provide an equal and opposite torque. The magnitude of this torque is given by TFf = (L/2) * Ff = (L/2) * (μs * FN).
Setting these two torque expressions equal, we have:
TD = TFf
(D/L) * Mg = (L/2) * (μs * FN)
(D/L) * Mg = (L/2) * (μs * (Mg/μs))
(D/L) * Mg = (L/2) * Mg
Here, we can observe that the mass M, acceleration due to gravity g, and coefficient of static friction μs cancel out.
Simplifying the equation, we get:
(D/L) = 1/2
Therefore, the minimum value of D/L is 1/2. In other words, D should be equal to half of L in order to keep the pole horizontal.
Since the coefficient of static friction μs is not involved in the final expression, there is no minimum value of μs required to keep the pole horizontal and prevent its end from sliding down the wall.
To determine the minimum value of μs, we need to consider the forces acting on the pole and find the conditions for equilibrium.
Let's break down the problem step by step:
1. Draw a free-body diagram of the pole:
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Label the point where the pole contacts the wall as A, the point where the rope is attached as B, and the center of mass of the pole as C.
2. Identify the forces acting on the pole:
- The weight of the pole Mg acting downward at C.
- The normal force FN acting perpendicular to the wall at A.
- The friction force Ffriction acting perpendicular and upward at A.
- Tension force FT acting horizontally from B to C.
3. Determine the torque equation:
Since the pole is in equilibrium, the net torque about point A must be zero. We can write:
Torque(A) = Torque(C)
Torque(A) = Mg * D (Clockwise torque due to the weight of the pole)
Torque(C) = FT * L/2 (Counter-clockwise torque due to the tension force)
Therefore, Mg * D = FT * L/2
4. Determine the vertical force equation:
Since the pole is in equilibrium, the net vertical force at A must be zero. We can write:
Vertical force(A) = FN - Mg + Ffriction = 0
FN = Mg - Ffriction
5. Determine the minimum value of μs:
The maximum friction force that can act at A without allowing the pole to slide down is μs * FN.
Therefore, we can write:
Ffriction ≤ μs * FN
Substituting FN = Mg - Ffriction:
Ffriction ≤ μs * (Mg - Ffriction)
Simplifying the equation, we have:
Ffriction ≤ μs * Mg - μs * Ffriction
(1 + μs) * Ffriction ≤ μs * Mg
Ffriction ≤ μs * Mg / (1 + μs)
Since Ffriction is positive, the maximum value of Ffriction occurs when equality holds:
Ffriction = μs * Mg / (1 + μs)
To keep the pole horizontal and not allow its end to slide down the wall, the vertical force of friction must be equal to the vertical component of the weight Mg.
Therefore, μs * Mg / (1 + μs) = Mg
Simplifying the equation, we have:
μs = Mg / (Mg - Mg)
μs = 1 / 0
Since division by zero is undefined, the minimum value of μs does not exist.
Therefore, there is no minimum value of μs that will keep the pole horizontal and not allow its end to slide down the wall. The pole will always slide down the wall under the given conditions.