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Chapter 3: Problem 75

(a) What is the smallest force parallel to a \(37^{\circ}\) incline needed to keep a \(100-\mathrm{N}\) weight from sliding down the incline if the coefficients of static and kinetic friction are both \(0.30 ?(b)\) What parallel force is required to keep the weight moving up the incline at constant speed? ( \(c\) ) If the parallel pushing force is \(94 \mathrm{~N}\), what will be the acceleration of the object? \((d)\) If the object in \((c)\) starts from rest, how far will it move in \(10 \mathrm{~s}\) ?

Short Answer

Expert verified
(a) 36.22 N, (b) 84.14 N, (c) 0.97 m/s虏, (d) 48.5 m

Step by step solution

01

Identify Known Values

Given an incline with an angle of \(37^{\circ}\), a weight of \(100\, \text{N}\), and coefficients of static and kinetic friction both \(0.30\). The gravitational acceleration \(g\) is \(9.8\, \text{m/s}^2\).
02

Calculate Normal Force

The normal force \(N\) is calculated as the component of the weight perpendicular to the incline. Use the formula:\[ N = W \cos(\theta) = 100\, \text{N} \times \cos(37^{\circ}) \approx 100\, \text{N} \times 0.7986 \approx 79.86\, \text{N} \]
03

Calculate Static Friction Force

The static friction force \(f_s\) is given by:\[ f_s = \mu_s \times N = 0.30 \times 79.86\, \text{N} \approx 23.96\, \text{N} \]
04

Determine Component of Weight Along Incline

The component of the weight along the incline is:\[ W_\parallel = W \sin(\theta) = 100\, \text{N} \times \sin(37^{\circ}) \approx 100\, \text{N} \times 0.6018 \approx 60.18\, \text{N} \]
05

Minimum Force to Prevent Sliding Down

To prevent the object from sliding down, the force \(F\) needs to balance the difference between the downhill force and the static friction:\[ F = W_\parallel - f_s \approx 60.18\, \text{N} - 23.96\, \text{N} \approx 36.22\, \text{N} \]
06

Calculate Kinetic Friction Force

The kinetic friction force \(f_k\) when moving is:\[ f_k = \mu_k \times N = 0.30 \times 79.86\, \text{N} \approx 23.96\, \text{N} \]
07

Force to Move Up at Constant Speed

To move up at constant speed, the applied force must overcome both the gravitational component and kinetic friction:\[ F' = W_\parallel + f_k \approx 60.18\, \text{N} + 23.96\, \text{N} \approx 84.14\, \text{N} \]
08

Acceleration with a 94 N Force

Apply Newton's second law along the incline:\[ F - W_\parallel - f_k = ma \]\[ 94\, \text{N} - 60.18\, \text{N} - 23.96\, \text{N} = ma \]\[ 9.86\, \text{N} = ma \]\[ a = \frac{9.86\, \text{N}}{10.20\, \text{kg}} \approx 0.97\, \text{m/s}^2 \]
09

Distance Traveled in 10s from Rest

Use the equation for motion from rest with constant acceleration:\[ s = ut + \frac{1}{2}at^2 \]Here, \(u = 0\, \text{m/s}, a = 0.97\, \text{m/s}^2, t = 10\, \text{s}\):\[ s = 0 + \frac{1}{2} \times 0.97\, \text{m/s}^2 \times (10\, \text{s})^2 = 48.5\, \text{m} \]

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Key Concepts

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

Friction
Friction is a force that opposes motion between two contacting surfaces. In physics, it plays a critical role in motion on inclined planes. There are two main types of friction forces relevant here: static and kinetic friction.
  • Static friction occurs when objects are stationary. It prevents them from sliding downwards.
  • Kinetic friction comes into play when objects are sliding.

Friction depends on the normal force, which is the force perpendicular to the contact surface, and the friction coefficient, which is a measure of how easily one surface moves over another. Static friction is calculated as \( f_s = \mu_s \times N \), where \( \mu_s \) is the coefficient of static friction and \( N \) is the normal force.
Similarly, kinetic friction is \( f_k = \mu_k \times N \). In our exercise, both static and kinetic friction coefficients are 0.30, meaning a moderate level of resistance to sliding.
Inclined plane
Inclined planes are flat surfaces tilted at an angle to horizontal. They're a foundational concept in physics as they illustrate the distribution of forces acting on an object. When a weight is placed on an incline, gravity can be resolved into two components:
  • One perpendicular (normal) to the incline, calculated with \( N = W \cos(\theta) \)
  • One parallel to the incline, found by \( W_\parallel = W \sin(\theta) \)

Forces such as friction and applied forces need to counterbalance these components to prevent motion or move an object, as shown in the given exercise with a \(37^{\circ}\) incline and weights moving up and down the incline.
Newton's laws
Newton's laws of motion are essential to solving problems involving forces. These include:
  • First law (inertia) states that an object remains at rest or moves at a constant speed in a straight line unless acted upon by a force.
  • Second law (force and acceleration) is relevant to our exercise: \( F = ma \), which links force, mass, and acceleration.
  • Third law (action-reaction) states that for every action, there is an equal and opposite reaction.

Newton's second law helps us determine the net force necessary to achieve or balance specific motions on an incline, as it incorporates gravitational, frictional, and applied forces. In our problem, it allowed us to calculate the acceleration when a force of 94 N acts on the weight moving up the incline.
Kinematics
Kinematics is the study of motion without considering forces causing the movement. In these problems, kinematics enables us to calculate how far an object travels when pushed with a certain force. When given an initial velocity, time, and acceleration, we use:
  • \( s = ut + \frac{1}{2}at^2 \) to find the distance \( s \).
  • Here, \( u \) is the initial velocity, \( a \) the acceleration and \( t \) the time.

In the given problem, if the object starts from rest (\( u = 0 \)) and has constant acceleration calculated from Newton's laws, we can determine the distance the object travels in a set time.

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Most popular questions from this chapter

A man who weighs \(1000 \mathrm{~N}\) on Earth stands on a scale on the surface of the mythical nonspinning planet Mongo. That body has a mass which is \(4.80\) times Earth's mass and a diameter which is \(0.500\) times Earth's diameter. Neglecting the effect of the Earth's spin, how much does the scale read?

A car coasting at \(20 \mathrm{~m} / \mathrm{s}\) along a horizontal road has its brakes suddenly applied and eventually comes to rest. What is the shortest distance in which it can be stopped if the friction coefficient between tires and road is \(0.90 ?\) Assume that all four wheels brake identically. If the brakes don't lock, the car stops via static friction. The friction force at one wheel, call it wheel 1 , is $$ F_{\mathrm{fl}}=\mu_{5} F_{N 1}=\mu_{s} F_{W 1} $$ where \(F_{W}\) is the weight carried by wheel 1 . We obtain the total friction force \(F_{\mathrm{f}}\) by adding such terms for all four wheels: $$ F_{\mathrm{f}}=\mu_{s} F_{W 1}+\mu_{s} F_{W_{2}}+\mu_{s} F_{W 3}+\mu_{s} F_{W 4}=\mu_{s}\left(F_{W_{1}}+F_{W 2}+F_{W 3}+F_{W_{4}}\right)=\mu_{s} F_{w} $$ where \(F_{W}\) is the total weight of the car. (Notice that we are assuming optimal braking at each wheel.) This friction force is the only unbalanced force on the car (we neglect air friction). Writing \(F=m a\) for the car with \(F\) replaced by \(-\mu_{s} F_{W}\) gives \(-\mu_{s} F_{W}=m a\), where \(m\) is the car's mass and the positive direction is taken as the direction of motion. However, \(F_{W}=m g\); so the car's acceleration is $$ a=-\frac{\mu_{s} F_{W}}{m}=-\frac{\mu_{s} m g}{m}=-\mu_{s} g=(-0.90)\left(9.81 \mathrm{~m} / \mathrm{s}^{2}\right)=-8.829 \mathrm{~m} / \mathrm{s}^{2} $$ We can determine how far the car went before stopping by solving the constant- \(a\) motion problem. Knowing that \(v_{i}=20 \mathrm{~m} / \mathrm{s}, v_{f}=0\), and \(a=-8.829 \mathrm{~m} / \mathrm{s}^{2}\), we find from \(v_{f}^{2}-v_{i}^{2}=2 a x\) that $$ x=\frac{(0-400) \mathrm{m}^{2} / \mathrm{s}^{2}}{-17.66 \mathrm{~m} / \mathrm{s}^{2}}=22.65 \mathrm{~m} \quad \text { or } \quad 23 \mathrm{~m} $$ If the four wheels had not all been braking optimally, the stopping distance would have been longer.

A car whose weight is \(F_{W}\) is on a ramp which makes an angle \(\theta\) to the horizontal. How large a perpendicular force must the ramp withstand if it is not to break under the car's weight? As rendered in Fig. \(3-6\), the car's weight is a force \(\overrightarrow{\mathbf{F}}_{W}\) that pulls straight down on the car. We take components of \(\overrightarrow{\mathbf{F}}\) along the incline and perpendicular to it. The ramp must balance the force component \(F_{W} \cos \theta\) if the car is not to crash through the ramp. In other words, the force exerted on the car by the ramp, upwardly perpendicular to the ramp, is \(F_{N}\) and \(F_{N}=F_{W} \cos \theta\).

Two forces act on a point object as follows: \(100 \mathrm{~N}\) at \(170.0^{\circ}\) and \(100 \mathrm{~N}\) at \(50.0^{\circ} .\) Find their resultant.

An object that weighs \(2700 \mathrm{~N}\) on the surface of the Earth is raised to a height (i.e., altitude) of two Earth radii above the surface. What will it weigh up there?
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