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Chapter 7: Problem 66

A combination lock has a 1.0 -cm-diameter knob that is part of the dial you turn to unlock the lock. To turn that knob, you grip it between your thumb and forefinger with a force of \(0.60 \mathrm{N}\) as you twist your wrist. Suppose the coefficient of static friction between the knob and your fingers is only 0.12 because some oil accidentally got onto the knob. What is the most torque you can exert on the knob without having it slip between your fingers?

Short Answer

Expert verified
The maximum torque that can be exerted on the knob without having it slip is \(0.00036 \mathrm{Nm}\).

Step by step solution

01

Identify the Known Variables

In the problem, it is given that the force exerted by the fingers, \(F\) is \(0.60 \mathrm{N}\), the radius of the knob, \(r\) is \(0.5 \mathrm{cm} = 0.005 \mathrm{m}\) (since radius is half the diameter), and the static coefficient of friction, \(渭_s\) is \(0.12\).
02

Calculate the Maximum Force of Static Friction

The maximum force of static friction, \(f_s\) can be calculated with the equation \(f_s = 渭_s 脳 F\). Substituting given values into the equation gives us \(f_s = 0.12 脳 0.60 = 0.072 \mathrm{N}\). This is the maximum force that can be applied without causing the knob to slip.
03

Calculate the Maximum Torque

Torque can be calculated using the formula \(蟿 = f_s 脳 r\). Substituting the calculated static friction and given radius into this equation gives us \(蟿 = 0.072 \mathrm{N} 脳 0.005 \mathrm{m} = 0.00036 \mathrm{Nm}\). Hence, this is the maximum torque that can be exerted on the knob without causing it to slip.

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Key Concepts

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

Understanding Static Friction
Static friction is the force that prevents surfaces from sliding past one another when they are in contact and at rest relative to each other. This is a force that must be overcome for motion to start. Imagine trying to push a piece of furniture across a room; the initial resistance you feel before the furniture starts moving is static friction at work.

In our example of the combination lock, static friction plays a crucial role. It prevents the knob from slipping through the fingers when a twisting force, or torque, is applied. When the fingers press against the knob, the static frictional force is what allows for the applied force to translate into a twisting motion, assuming that this force does not exceed the maximum force that static friction can sustain before 'breaking free' and allowing slip.
Force Exertion and Its Application
The term force exertion refers to the application of force on an object. The interesting thing about exerting force is that it's not just about how much force is applied, but also how this force is applied. For instance, when you twist a knob, you're applying force in a circular path, which is quite different from pushing it in a straight line.

In the case of the lock knob, exerting a force of 0.60 N through a static-frictional contact allows a torque to be generated. Torque is the rotational equivalent of linear force and is important when you're dealing with rotational motion. To exert the most effective torque, you need to maximize the use of friction before the knob slips, which leads us to the interplay of force exertion and friction.
Coefficient of Static Friction
The coefficient of static friction, often denoted by the Greek letter \(渭_s\), is a value that describes how much frictional force can be generated between two surfaces without motion. It is a unitless value that represents the ratio of the maximum static frictional force to the normal force鈥攖his is essentially the force pressing the two surfaces together.

In the context of the combination lock, a \(渭_s\) of 0.12 signifies a low friction scenario, potentially due to an oily surface. Despite that, it does not eliminate friction altogether but instead tells you that the force your fingers can apply is limited. To calculate the torque effectively, you would need to take into account this coefficient, along with the exerted force and the distance from the center of the knob (its radius), which all combine to give you the amount of torque you can apply before the slip occurs.

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Most popular questions from this chapter

A \(1.0 \mathrm{kg}\) ball and a \(2.0 \mathrm{kg}\) ball are connected by a \(1.0-\mathrm{m}\) -long rigid, massless rod. The rod and balls are rotating clockwise about their center of gravity at 20 rpm. What torque will bring the balls to a halt in \(5.0 \mathrm{s} ?\)

On page 224 there is a photograph of a girl pushing on a large stone sphere. The sphere has a mass of \(8200 \mathrm{kg}\) and a radius of \(90 \mathrm{cm} .\) Suppose that she pushes on the sphere tangent to its surface with a steady force of \(50 \mathrm{N}\) and that the pressured water provides a frictionless support. How long will it take her to rotate the sphere one time, starting from rest?

The bunchberry flower has the fastest-moving parts ever seen in a plant. Initially, the stamens are held by the petals in a bent position, storing energy like a coiled spring. As the petals release, the tips of the stamens fly up and quickly release a burst of pollen. Figure \(\mathrm{P} 7.78\) shows the details of the motion. The tips of the stamens act like a catapult, flipping through a \(60^{\circ}\) angle; the times on the earlier photos show that this happens in just 0.30 ms. We can model a stamen tip as a 1.0 -mm- long, \(10 \mu\) g rigid rod with a \(10 \mu g\) anther sac at one end and a pivot point at the opposite end. Though an oversimplification, we will model the motion by assuming the angular acceleration is constant throughout the motion. How large is the "straightening torque"? (You can omit gravitational forces from your calculation; the gravitational torque is much less than this.) A. \(2.3 \times 10^{-7} \mathrm{N} \cdot \mathrm{m}\) B. \(3.1 \times 10^{-7} \mathrm{N} \cdot \mathrm{m}\) C. \(2.3 \times 10^{-5} \mathrm{N} \cdot \mathrm{m}\) D. \(3.1 \times 10^{-5} \mathrm{N} \cdot \mathrm{m}\)

Hold your arm outstretched so that it is horizontal. Estimate the mass of your arm and the position of its center of gravity. What is the gravitational torque on your arm in this position, computed around the shoulder joint?

What is the angular position in radians of the minute hand of a clock at (a) 5: 00 , (b) \(7: 15,\) and (c) \(3: 35 ?\)
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