/*! 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 99 An \(11 \mathrm{~kg}\) block of ... [FREE SOLUTION] | 91影视

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Chapter 6: Problem 99

An \(11 \mathrm{~kg}\) block of steel is at rest on a horizontal table. The coefficient of static friction between block and table is \(0.52 .\) (a) What is the magnitude of the horizontal force that will put the block on the verge of moving? (b) What is the magnitude of a force acting upward \(60^{\circ}\) from the horizontal that will put the block on the verge of moving? (c) If the force acts downward at \(60^{\circ}\) from the horizontal, how large can its magnitude be without causing the block to move?

Short Answer

Expert verified
(a) 56.056 N, (b) 62.34 N, (c) 161.12 N.

Step by step solution

01

Understanding the problem

We are given a block of mass 11 kg resting on a table with a coefficient of static friction \( \mu_s = 0.52 \). We need to find three forces: (a) a horizontal force, (b) an upward 60-degree force, and (c) a downward 60-degree force that will cause the block to start moving.
02

Calculating the normal force

The normal force \( N \) for a block resting on a horizontal surface is equal to the weight of the block because there are no vertical forces acting on it besides gravity. Thus, \( N = mg = 11 \, \text{kg} \times 9.8 \, \text{m/s}^2 = 107.8 \, \text{N} \).
03

Solving for horizontal force (a)

The maximum static friction force \( F_{s} \) that can oppose the movement is given by \( F_{s} = \mu_s N \). Substituting the values, we find \( F_{s} = 0.52 \times 107.8 = 56.056 \, \text{N} \). This is the magnitude of the horizontal force needed to move the block.
04

Solving for upward 60-degree force (b)

When the force is applied at a 60-degree angle upwards, the vertical component reduces the normal force. The normal force is now \( N = mg - F \sin 60^\circ \). For static equilibrium, \( F \cos 60^\circ = \mu_s N \). Solve for \( F \) using the equations: \( \mu_s (mg - F \sin 60^\circ) = F \cos 60^\circ \), leading to \( F = \frac{\mu_s mg}{\cos 60^\circ + \mu_s \sin 60^\circ} = \frac{0.52 \times 107.8}{0.5 + 0.52 \times 0.866} \approx 62.34 \, \text{N} \).
05

Solving for downward 60-degree force (c)

When the force is applied at a 60-degree angle downwards, the vertical component increases the normal force. The normal force is now \( N = mg + F \sin 60^\circ \). For static equilibrium, \( F \cos 60^\circ = \mu_s N \). Solve for \( F \) using the equations: \( \mu_s (mg + F \sin 60^\circ) = F \cos 60^\circ \), leading to \( F = \frac{\mu_s mg}{\cos 60^\circ - \mu_s \sin 60^\circ} = \frac{0.52 \times 107.8}{0.5 - 0.52 \times 0.866} \approx 161.12 \, \text{N} \).

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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 a force that keeps an object at rest when it's placed on a surface. It acts between two surfaces that are in contact and prevents sliding. In our scenario, the 11 kg block of steel sits on the table without moving because of static friction.
The static friction force has a maximum value, which means there's a limit to how much friction can resist before the object begins to move. The formula for this maximum static friction is:
  • \( F_{s} = \mu_s N \)
Here, \( \mu_s \) represents the coefficient of static friction, a number describing how rough or smooth the surfaces are. A higher value means more friction.
In the exercise, the maximum static friction (56.056 N) tells us the strongest horizontal force that can be applied before the block starts moving. Think of it as the final "grip" the table has on the block before it slides.
Normal Force
Normal force is the support force exerted upon an object that is in contact with another stable object. It acts perpendicular to the surface. For the steel block at rest on the table, the normal force is equal to the weight of the block.
The equation for normal force when no other vertical forces are acting is:
  • \( N = mg \)
where \( m \) is mass in kilograms and \( g \) is the acceleration due to gravity (\( 9.8 \, \text{m/s}^2 \)).
In parts (b) and (c) of the problem, the angle of applied force changes the normal force. For a force acting upward, the normal force decreases because part of that force counteracts gravity. Conversely, applying force downward adds to gravity, increasing the normal force.
This change influences how strong the static friction can be before movement starts, as a smaller normal force limits friction's resistance, making it easier for the block to move.
Equilibrium
Equilibrium refers to a state where all forces acting on an object are balanced, resulting in no movement. The block remains stationary due to the balance between gravitational force, normal force, and static friction.
This concept is crucial when analyzing forces in motion problems. For the block to remain in equilibrium, the net force acting on it must be zero. This principle helps us determine the magnitude of applied forces necessary to initiate movement.
When forces are applied at an angle, such as in our exercise, the concept of equilibrium guides the calculations:
  • For an upward force at a \(60^{\circ}\) angle, equilibrium equates the horizontal component of the force with the maximum static friction, factoring in adjustments to the normal force.
  • Similarly, for a downward angle, equilibrium also takes into account the increased normal force when calculating static resistance.
Maintaining equilibrium is about finding the point just before motion starts, balancing all forces precisely to the stick-slip threshold.

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Most popular questions from this chapter

A child places a picnic basket on the outer rim of a merrygo-round that has a radius of \(4.6 \mathrm{~m}\) and revolves once every \(30 \mathrm{~s}\). (a) What is the speed of a point on that rim? (b) What is the lowest value of the coefficient of static friction between basket and merry-go-round that allows the basket to stay on the ride?

A ski that is placed on snow will stick to the snow. However, when the ski is moved along the snow, the rubbing warms and partially melts the snow, reducing the coefficient of kinetic friction and promoting sliding. Waxing the ski makes it water repellent and reduces friction with the resulting layer of water. A magazine reports that a new type of plastic ski is especially water repellent and that, on a gentle \(200 \mathrm{~m}\) slope in the Alps, a skier reduced his top-to-bottom time from \(61 \mathrm{~s}\) with standard skis to \(42 \mathrm{~s}\) with the new skis. Determine the magnitude of his average acceleration with (a) the standard skis and (b) the new skis. Assuming a \(3.0^{\circ}\) slope, compute the coefficient of kinetic friction for (c) the standard skis and (d) the new skis.

A student wants to determine the coefficients of static friction and kinetic friction between a box and a plank. She places the box on the plank and gradually raises one end of the plank. When the angle of inclination with the horizontal reaches \(30^{\circ}\), the box starts to slip, and it then slides \(2.5 \mathrm{~m}\) down the plank in \(4.0 \mathrm{~s}\) at constant acceleration. What are (a) the coefficient of static friction and (b) the coefficient of kinetic friction between the box and the plank?

The floor of a railroad flatcar is loaded with loose crates having a coefficient of static friction of \(0.25\) with the floor. If the train is initially moving at a speed of \(48 \mathrm{~km} / \mathrm{h}\), in how short a distance can the train be stopped at constant acceleration without causing the crates to slide over the floor?

A house is built on the top of a hill with a nearby slope at angle \(\theta=45^{\circ}\) (Fig. 6-55). An engineering study indicates that the slope angle should be reduced because the top layers of soil along the slope might slip past the lower layers. If the coefficient of static friction between two such layers is \(0.5\), what is the least angle \(\phi\) through which the present slope should be reduced to prevent slippage?
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