/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 144 A toy cart of mass \(m_{1}\) mov... [FREE SOLUTION] | 91影视

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Chapter 4: Problem 144

A toy cart of mass \(m_{1}\) moves on frictionless wheels as it is pulled by a string under tension \(T .\) A block of mass \(m_{2}\) rests on top of the cart. The coefficient of static friction between the cart and the block is \(\mu .\) Find the maximum tension \(T\) that will not cause the block to slide on the cart if the cart rolls on (a) a horizontal surface; (b) up a ramp of angle \(\theta\) above the horizontal. In both cases, the string is parallel to the surface on which the cart rolls.

Short Answer

Expert verified
Answer: The maximum tensions in the given scenarios are (a) on a horizontal surface - \(T_{max} = \mu m_{2}g\) and (b) on an inclined ramp at an angle 胃 - \(T_{max} = m_{2}g\sin(\theta) + \mu m_{2}g\cos(\theta)\).

Step by step solution

01

Analyze the forces on the block

Since the block is at rest on the cart, the force of gravity (\(m_{2}g\)) acts downward, and the static friction force between the block and the cart (\(f_{s}\)) acts in the opposite direction to the force exerted by the tension \(T\). The normal force also acts on the block, but as there is no vertical motion, it is equal to the gravitational force acting on the block.
02

Set up the equations for the forces

As the block is in equilibrium, the net force acting on it is zero. This can be written as: \(f_{s} = m_{2}a_{2}\), where \(a_{2}\) is the acceleration of the block.
03

Calculate the maximum static friction force

The maximum static friction force that can act on the block before it starts sliding is given by \(f_{s\_max} = \mu N\), where \(\mu\) is the coefficient of static friction and \(N\) is the normal force. Since \(N = m_{2}g\), we have \(f_{s\_max} = \mu m_{2}g\).
04

Find the maximum tension

For the block to not slide on the cart, the tension force must not exceed the maximum static friction force; therefore, we have \(T \leq f_{s\_max}\). This means that the maximum tension in this scenario is \(T_{max} = \mu m_{2}g\). ##Inclined Ramp##
05

Analyze the forces on the block for an inclined ramp

Now, the cart is on an inclined ramp at an angle \(\theta\). The gravitational force \(m_{2}g\) has two components: \(m_{2}g\cos(\theta)\), perpendicular to the ramp, and \(m_{2}g\sin(\theta)\), parallel to the ramp, down the incline. We also have the normal force \(N\) and the static friction force \(f_s\).
06

Set up the equations for the forces on the inclined ramp

For the block to be in equilibrium and not slide on the cart, \(T = m_{2}g\sin(\theta) + f_{s}\).
07

Calculate the maximum static friction force on the inclined ramp

The maximum static friction force can be found as \(f_{s\_max} = \mu N\). In this case, since \(N = m_{2}g\cos(\theta)\), we have \(f_{s\_max} = \mu m_{2}g\cos(\theta)\).
08

Find the maximum tension for the inclined ramp

For the block to not slide on the cart, the tension force must not exceed the maximum static friction force, just like in the case of a horizontal surface. Therefore, we must have \(T \leq m_{2}g\sin(\theta) + f_{s\_max}\). Thus, the maximum tension in this scenario is \(T_{max} = m_{2}g\sin(\theta) + \mu m_{2}g\cos(\theta)\).

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Most popular questions from this chapter

A roller coaster is towed up an incline at a steady speed of $0.50 \mathrm{m} / \mathrm{s}$ by a chain parallel to the surface of the incline. The slope is \(3.0 \%,\) which means that the elevation increases by \(3.0 \mathrm{m}\) for every \(100.0 \mathrm{m}\) of horizontal distance. The mass of the roller coaster is \(400.0 \mathrm{kg}\). Ignoring friction, what is the magnitude of the force exerted on the roller coaster by the chain?
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