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Chapter 12: Problem 9

If you take the pivot point at the application point of one force in a static- equilibrium problem, that force doesn't enter the torque equation. Does that make the force irrelevant to the problem? Explain.

Short Answer

Expert verified
No, taking the pivot point at the application point of a force does not make the force irrelevant. Although it doesn't contribute to the torque in this case, the force can still impact the overall mechanical behavior of the system, such as causing a linear motion or affecting the equilibrium status.

Step by step solution

01

Understand the concept of torque

Torque, denoted by the Greek letter tau \(\tau\), is the rotational equivalent of linear force. It is calculated as the product of the force and the distance from the pivot point at which it is applied, multiplied by the sine of the angle between the force vector and the lever arm vector.
02

Connection between pivot point and force

When we consider a pivot point at the same location where the force is applied, the distance from the pivot point becomes zero. Since torque is calculated by the product of force and distance, if the distance is zero, then the torque will also be zero.
03

Relevance of the force

Although taking the pivot point at the application point of the force results in zero torque for that force, it doesn't render the force irrelevant to the problem. The force can still be producing linear motion or changing the equilibrium status of the system. It therefore contributes to the overall mechanical behavior of the system, which could involve both translational and rotational motion.

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Most popular questions from this chapter

A 5.0 -m-long ladder has mass \(9.5 \mathrm{kg}\) and is leaning against a frictionless wall, making a \(66^{\circ}\) angle with the horizontal. If the coefficient of friction between ladder and ground is \(0.42,\) what's the mass of the heaviest person who can safely ascend to the top of the ladder? (The center of mass of the ladder is at its center.)

A uniform solid cone of height \(h\) and base diameter \(\frac{1}{3} h\) sits on the board of Fig. \(12.27 .\) The coefficient of static friction between the cone and incline is \(0.63 .\) As the slope of the board is increased, will the cone first tip over or first begin sliding? (Hint: Start with an integration to find the center of mass.)

A uniform \(5.0-\mathrm{kg}\) ladder is leaning against a frictionless vertical wall, with which it makes a \(15^{\circ}\) angle. The coefficient of friction between ladder and ground is \(0.26 .\) Can a \(65-\mathrm{kg}\) person climb to the top of the ladder without it slipping? If not, how high can that person climb? If so, how massive a person would make the ladder slip?

What horizontal force applied at its highest point is necessary to keep a wheel of mass \(M\) from rolling down a slope inclined at angle \(\theta\) to the horizontal?

You're investigating ladder safety for the Consumer Product Safety Commission. Your test case is a uniform ladder of mass \(m\) leaning against a frictionless vertical wall with which it makes an angle \(\theta .\) The coefficient of static friction at the floor is \(\mu .\) Your job is to find an expression for the maximum mass of a person who can climb to the top of the ladder without its slipping. With that result, you're to show that anyone can climb to the top if \(\mu \geq \tan \theta\) but that no one can if \(\mu<\frac{1}{2} \tan \theta.\)
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