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Chapter 8: Problem 89

A lever AB of negligible weight is loosely fitted onto a 2.5-in.-diameter fixed shaft. Knowing that the coefficient of static friction between the fixed shaft and the lever is 0.15, determine the force P required to start the lever rotating clockwise.

Short Answer

Expert verified
Calculate \( P \) using the torque equation: \( P = \frac{0.15 \cdot N \cdot 1.25}{L} \).

Step by step solution

01

Understand the Problem

We have a lever AB on a fixed shaft with a coefficient of static friction \( \mu_s = 0.15 \). The task is to determine the force \( P \) required to overcome the static friction and start the lever rotating clockwise.
02

Calculate Normal Force

Assuming the lever is horizontal and not influenced by other vertical forces (due to negligible weight), the normal force \( N \) exerted by the lever on the shaft is horizontal, counteracting the applied force \( P \).
03

Determine Frictional Force

The frictional force \( F_f \) that needs to be overcome to initiate movement is given by the equation \( F_f = \mu_s \cdot N \). In this situation, \( F_f = 0.15 \cdot N \).
04

Relate Frictional Force to Torque

To start the lever moving, the applied force \( P \) must create a torque greater than the frictional torque resisting motion. The frictional torque \( T_f \) is \( T_f = F_f \cdot r \), where \( r = \frac{2.5}{2} = 1.25 \) inches is the radius of the shaft.
05

Calculate Required Force P

To find \( P \), assume it acts at the end of the lever, perpendicular to the line from the shaft to the point where \( P \) is applied. The torque created by \( P \) is \( T_P = P \cdot L \), where \( L \) is the distance from the shaft axis to the point of application. Set \( T_P = T_f \) and solve for \( P \).
06

Solve the Torque Equation

We have \( P \cdot L = (0.15 \cdot N) \cdot 1.25 \). If we solve this equation for \( P \), we get \( P = \frac{0.15 \cdot N \cdot 1.25}{L} \). Insert any given values for \( L \) in the problem if available to calculate \( P \).

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Key Concepts

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

Lever Mechanics
Lever mechanics is a fundamental concept in physics, focusing on how levers provide mechanical advantage in lifting or moving loads. A lever consists of a rigid bar pivoting around a fixed point, known as the fulcrum.
In the exercise, the lever AB pivots around a fixed shaft. Understanding how levers work is crucial because they can amplify force, allowing a small force to move a larger load.
There are three classes of levers based on the positioning of the load, effort (force), and fulcrum:
  • First-class lever: The fulcrum is between the force and the load (e.g., a seesaw).
  • Second-class lever: The load is between the fulcrum and the force (e.g., a wheelbarrow).
  • Third-class lever: The force is between the fulcrum and the load (e.g., a pair of tweezers).
In the given problem, the lever likely functions similarly to a lever seeking balance, where the force applied must overcome static friction to rotate around the shaft.
Torque Calculation
Torque is a measure of the rotational force applied to an object. In lever mechanics, torque depends on the force applied and how far it is applied from the axis of rotation or pivot point.
Calculated as a product of force and distance from the fulcrum:\[ \text{Torque} = \text{Force} \times \text{Distance} \]In this context, two torques are involved:
  • Frictional torque: The torque resisting the lever's motion due to static friction. This torque equals the frictional force multiplied by the radius of the shaft.
  • Applied torque: The torque generated by the force P applied at some distance L from the shaft.
To start the lever rotating, the applied torque must exceed the frictional torque. This means the force applied at the specified distance must overcome the resistance from the static friction around the shaft.
Frictional Force
Frictional force is a force that opposes the relative motion of two surfaces in contact. When considering static friction, it resists the initial motion and must be overcome to start moving.
In our problem, the lever is resting on a fixed shaft, creating friction at the contact points. The static friction force \( F_f \) is crucial, calculated with the formula:\[ F_f = \mu_s \times N \]where \( \mu_s \) is the static friction coefficient (0.15 in this scenario) and \( N \) is the normal force exerted by the lever on the shaft.
This frictional force directly relates to the frictional torque, resisting the lever's rotation. The students need to grasp that overcoming this frictional force requires calculating the exact force P that generates enough torque to outmatch this frictional resistance.

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Most popular questions from this chapter

A loaded railroad car has a mass of \(30 \mathrm{Mg}\) and is supported by eight 800 -mm-diameter wheels with 125 -mm-diameter axles. Knowing that the coefficients of friction are \(\mu_{s}=0.020\) and \(\mu_{k}=0.015\) determine the horizontal frce required (a) to start the car moving, (b) to keep the car moning at a constant speed. Neglect rolling resistance between the wheels and the rails.

A 120-lb cabinet is mounted on casters that can be locked to prevent their rotation. The coefficient of static friction between the floor and each caster is 0.30. Assuming that the casters at both A and B are locked, determine (a) the force P required to move the cabinet to the right, (b) the largest allowable value of h if the cabinet is not to tip over.

A rope having a weight per unit length of \(0.4 \mathrm{lb} / \mathrm{ft}\) is wound \(2 \frac{1}{2}\) times around a horizontal rod. Knowing that the coefficient of static friction between the rope and the rod is \(0.30,\) determine the minimum length \(x\) of rope that should be left hanging if a \(100-1 b\) load is to be supported.

Two slender rods of negligible weight are pin-connected at \(C\) and attached to blocks \(A\) and \(B\), each with a weight \(W .\) Knowing that \(\theta=80^{\circ}\) and that the coefficient of static friction between the blocks and the horizontal surface is 0.30 , determine the largest value of \(P\) for which equilibrium is maintained.

A \(12^{\circ}\) wedge is used to spread a split ring. The coefficient of static friction between the wedge and the ring is \(0.30 .\) Knowing that a force \(P\) with a magnitude of \(120 \mathrm{N}\) was required to insert the wedge, determine the magnitude of the forces exerted on the ring by the wedge after insertion.
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