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Chapter 11: Problem 60

The left-hand end of a slender uniform rod of mass \(m\) is placed against a vertical wall. The rod is held in a horizontal position by friction at the wall and by a light wire that runs from the right-hand end of the rod to a point on the wall above the rod. The wire makes an angle \(\theta\) with the rod. (a) What must the magnitude of the friction force be in order for the rod to remain at rest? (b) If the coefficient of static friction between the rod and the wall is \(\mu_{\mathrm{s}},\) what is the maximum angle between the wire and the rod at which the rod doesn't slip at the wall?

Short Answer

Expert verified
a. The magnitude of the friction force to keep the rod at rest is \(mg\). b. The maximum angle between the wire and the rod at which the rod doesn't slip is given by \(\theta = \tan^{-1}(\mu_{\mathrm{s}})\).

Step by step solution

01

Identify Forces Acting on the Rod

Three forces are acting on the rod: the weight acting downward at the rod's center of mass that is \(mg\), the friction force acting upward at the wall that keeps the rod from falling, and the tension in the wire that can be divided into two components, one perpendicular to the rod \(T\sin \theta\) and the other parallel to the rod \(T\cos \theta\). The rod is in equilibrium, so all forces in the vertical and horizontal direction must balance out.
02

Calculate Friction Force

Consider the vertical forces and set up the equation from Newton's second law, \(\Sigma F_y = 0 \rightarrow friction = mg - T\sin \theta\). We need to know the tension in the wire in order to solve the equation. To find the tension, consider the horizontal forces, \(\Sigma F_x = 0 \rightarrow T\cos \theta = friction\). Isolate \(T\) from the second equation and substitute into the first equation to find the friction force.
03

Compute Maximum Angle

In order not to slip, the required frictional force must be less than or equal to the maximum static friction, \(\mu_{\mathrm{s}}N\), where \(N\) is normal force. Here, the normal force equals to the horizontal component of tension, \(T\cos \theta\). Equate the frictional force with the maximum static friction, and solve for \(\theta\) to find the maximum angle at which the rod doesn't slip.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Friction Force
In the context of static equilibrium, the friction force is a critical component that keeps objects in place when other forces try to cause movement. Here, the friction force acts between the slender rod and the vertical wall to prevent the rod from slipping downwards. This force counteracts the gravitational pull on the rod's mass, ensuring stability.

Friction force is determined by two main factors:
  • The surface texture or roughness between the two interacting surfaces, which in this case, is the wall and the rod.
  • The magnitude of the normal force pressing the two surfaces together.
The direction of the friction force is opposite to the direction of the potential movement, thus acting vertically upwards to support the rod.

In mathematical terms, the friction force can be calculated using the equation derived from balancing the forces in the vertical direction: \( ext{friction force} = mg - T\sin \theta \). Here, \(mg\) represents the weight of the rod acting downwards, and \(T\sin \theta\) is the vertical component of the tension in the wire, acting upwards.
Coefficient of Static Friction
The coefficient of static friction, often denoted as \(\mu_s\), is a dimensionless quantity that represents the amount of frictional resistance one surface has when compared against another, in the absence of movement. It is an essential aspect in determining whether the rod remains stationary against the wall.

This coefficient varies depending on the materials of the interacting surfaces, and can be typically found through empirical measurement or reference tables. In our problem, it is used to determine the capability of the friction force to prevent slipping as the angle of the wire changes. The static friction force can be expressed as:

\[ f_s \leq \mu_s N \]

Where \(f_s\) is the static friction force and \(N\) is the normal force. In our exercise, the normal force is equivalent to the horizontal component of the tension in the wire, \(T\cos \theta\). This helps apply the condition required for the rod to remain in static equilibrium.
Tension in the Wire
Tension in the wire is a crucial force that prevents the rod from rotating or falling. It acts as a stabilizing force pulling the right-hand end of the rod towards the point of attachment on the wall. Tension can be resolved into two components:
  • \(T\sin \theta\): This vertical component supports the rod against gravity, complementing the friction force.
  • \(T\cos \theta\): This horizontal component presses the rod against the wall and influences the normal force.


The correct level of tension in the wire is essential for maintaining static equilibrium. We can calculate this by considering the equilibrium of forces in both horizontal and vertical directions, as depicted by the equations:
  • \(\Sigma F_y = 0: mg - T\sin \theta + \,\text{friction} = 0\)
  • \(\Sigma F_x = 0: T\cos \theta - \,\text{normal force} = 0\)
By solving these equations, you determine the necessary tension to achieve a balanced and stable system.
Equilibrium Forces
Equilibrium forces are beneficial when trying to understand how a system remains in a stable state. In the case of our rod, it is under static equilibrium, which means that the sum of forces and moments acting upon it are zero, allowing it to remain motionless.

Static equilibrium can be split into two categories:
  • Translational Equilibrium: Refers to the absence of any resultant force that causes linear movement. It is represented by the conditions \(\Sigma F_x = 0\) and \(\Sigma F_y = 0\).
  • Rotational Equilibrium: Refers to the absence of any resultant torque that causes rotational movement. Here, rotational balance ensures that moments about any point within the system sum up to zero.
In the original exercise, this means that:
  • All vertical forces including the weight of the rod and vertical tension component must balance each other.
  • All horizontal forces, involving the normal force and horizontal tension component, counteract each other.
By considering these constraints, we can ensure that the net effect of all forces on the system is zero, firmly holding the rod in its equilibrium position.

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Most popular questions from this chapter

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