/*! 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 102 A wood block of mass \(m\) rests... [FREE SOLUTION] | 91Ó°ÊÓ

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A wood block of mass \(m\) rests on a larger wood block of mass \(M\) that rests on a wooden table. The coefficients of static and kinetic friction between all surfaces are \(\mu_{\mathrm{s}}\) and \(\mu_{\mathrm{k}}\), respectively. What is the minimum horizontal force, \(F\), applied to the lower block that will cause it to slide out from under the upper block?

Short Answer

Expert verified
The minimum force is \( F = \mu_s \cdot m \cdot g + \mu_k \cdot (m + M) \cdot g \).

Step by step solution

01

Understand the Forces Involved

First, identify the forces acting on the blocks. The larger block with mass \(M\) has a force \(F\) applied to it. Both blocks experience frictional forces due to their contact with each other and the table. The smaller block with mass \(m\) is resting on top of the larger block.
02

Calculate Maximum Static Friction

The maximum static friction between the two blocks, which prevents the smaller block from slipping off the larger block, is given by \(f_{s, ext{max}} = \mu_s \cdot m \cdot g\), where \(g\) is the acceleration due to gravity. This frictional force is what needs to be overcome to cause slipping.
03

Determine Net Force for Movement

For the larger block to move under the smaller block without causing it to slip off, the applied force \(F\) must be greater than the maximum static frictional force. So, we require \(F > f_{s, ext{max}} = \mu_s \cdot m \cdot g\).
04

Consider the Friction on Table

The friction between the larger block and the table must also be considered. The kinetic friction here is \(f_{k} = \mu_k \cdot (m + M) \cdot g\). This means the applied force must not only overcome the static friction between the blocks but also the kinetic friction with the table.
05

Establish the Minimum Required Force

Thus, the minimum force \(F\) which causes the larger block \(M\) to slide is given by equating it to the sum of the forces, therefore \(F = \mu_s \cdot m \cdot g + \mu_k \cdot (m + M) \cdot g\). This expression accounts for both static friction above and kinetic friction below.

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Key Concepts

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

Static Friction
Static friction is the force that keeps objects at rest when they are in contact with each other. It is the frictional force that must be overcome to initiate motion between stationary objects. In our example, the smaller wood block sits on top of the larger block, and there is a static frictional force between them. This static friction prevents the smaller block from sliding off as the larger block begins to move. The maximum static frictional force, before slipping occurs, can be calculated using the formula:
  • \[ f_{s, \text{max}} = \mu_s \cdot m \cdot g \]
Here, \(\mu_s\) represents the coefficient of static friction, \(m\) is the mass of the smaller block, and \(g\) is the gravitational acceleration. The coefficient \(\mu_s\) depends on the materials of the surfaces in contact. Static friction is crucial as it represents the threshold force required to overcome rest inertia in the smaller block.
Kinetic Friction
Once an object begins to move, static friction is replaced by kinetic friction. Kinetic friction is typically lower than static friction. It acts in the opposite direction to the movement, trying to resist the relative motion between surfaces. For the wooden blocks, once the lower block with mass \(M\) starts sliding on the table, kinetic friction becomes significant. It is calculated as follows:
  • \[ f_k = \mu_k \cdot (m + M) \cdot g \]
Here, \(\mu_k\) is the coefficient of kinetic friction. The formula accounts for the gravitational pull on both blocks combined, \((m + M)\). Successfully understanding kinetic friction is key to determining how a moving object will continue to behave on various surfaces. It highlights the transition from a static scenario to one involving movement and explains why less force is needed to keep an object moving compared to starting its motion.
Force Calculations
Force calculations help determine the necessary forces to instigate and maintain motion between objects. In this context, to move the larger block without displacing the upper block, one must calculate the minimum horizontal force \(F\) needed. Firstly, the force \(F\) needs to surpass the maximum static friction to initiate movement within the blocks. Secondly, \(F\) also has to overcome the kinetic friction between the larger block and the tabletop.The minimum force required is the sum of these frictions:
  • \[ F = \mu_s \cdot m \cdot g + \mu_k \cdot (m + M) \cdot g \]
This equation embodies both forces in play: overcoming initial static friction and sustaining movement with kinetic friction. Understanding this calculation provides insight into how different factors, like friction coefficients and masses, influence the force needed for movement. Accurate force calculation is essential in predicting and controlling motion in physical systems.

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Most popular questions from this chapter

A 0.16-g spider hangs from the middle of the first thread of its future web. The thread makes an angle of \(7.2^{\circ}\) with the horizontal on both sides of the spider. (a) What is the tension in the thread? (b) If the angle made by the thread had been less than \(7.2^{\circ}\), would its tension have been greater than, less than, or the same as in part (a)? Explain.

Playing a Violin The tension in a violin string is \(2.7 \mathrm{N}\). When pushed down against the neck of the violin, the string makes an angle of \(4.1^{\circ}\) with the horizontal. (a) With what force must you push down on the string to bring it into contact with the neck? (b) If the angle were less than \(4.1^{\circ}\), would the required force be greater than, less than, or the same as in part (a)? Explain.

Hopping into your Porsche, you floor it and accelerate at \(12 \mathrm{m} / \mathrm{s}^{2}\) without spinning the tires. Determine the minimum coefficient of static friction between the tires and the road needed to make this possible.

A \(110-\mathrm{kg}\) box is loaded into the trunk of a car. If the height of the car's bumper decreases by \(13 \mathrm{cm},\) what is the force constant of its rear suspension?

Two drivers traveling side-by-side at the same speed suddenly see a deer in the road ahead of them and begin braking. Driver 1 stops by locking up his brakes and screeching to a halt; driver 2 stops by applying her brakes just to the verge of locking, so that the wheels continue to turn until her car comes to a complete stop. (a) All other factors being equal, is the stopping distance of driver 1 greater than, less than, or equal to the stopping distance of driver \(2 ?\) (b) Choose the best explanation from among the following: I. Locking up the brakes gives the greatest possible braking force. II. The same tires on the same road result in the same force of friction. III. Locked-up brakes lead to sliding (kinetic) friction, which is less than rolling (static) friction.

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