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Chapter 3: Problem 57

A 20 -kg crate hangs at the end of a long rope. Find its acceleration (magnitude and direction) when the tension in the rope is ( a) 250 \(\mathrm{N},(b) 150 \mathrm{~N},(c)\) zero, (d) \(196 \mathrm{~N}\).

Short Answer

Expert verified
(a) 2.7 m/s虏 up, (b) 2.3 m/s虏 down, (c) 9.8 m/s虏 down, (d) 0 m/s虏.

Step by step solution

01

Understand the Forces

The crate is subject to two main forces: gravity, which pulls it downward, and the tension in the rope, which can pull it upward or leave it in free fall if there is no tension. The weight of the crate, due to gravity, is calculated as \[ F_g = m \cdot g = 20 \times 9.8 = 196 \, \text{N}, \]where \( m = 20 \, \text{kg} \) is the mass and \( g = 9.8 \, \text{m/s}^2 \) is the acceleration due to gravity.
02

Apply Newton's Second Law

Newton's second law states that \( F_{net} = m \cdot a \), where \( F_{net} \) is the net force acting on the object, \( m \) is its mass, and \( a \) is its acceleration. Here, the net force \( F_{net} \) is the difference between the tension \( T \) in the rope and the gravitational force \( F_g \). This gives us: \[ F_{net} = T - F_g = T - 196. \]This means the acceleration \( a \) can be calculated from:\[ a = \frac{T - 196}{20}. \]
03

Solve for Each Condition

Now, solve for each tension value given:(a) For \( T = 250 \, \text{N} \):\[ a = \frac{250 - 196}{20} = \frac{54}{20} = 2.7 \, \text{m/s}^2. \]The crate accelerates upward at \( 2.7 \, \text{m/s}^2 \).(b) For \( T = 150 \, \text{N} \):\[ a = \frac{150 - 196}{20} = \frac{-46}{20} = -2.3 \, \text{m/s}^2. \]The crate accelerates downward at \( 2.3 \, \text{m/s}^2 \).(c) For \( T = 0 \, \text{N} \):\[ a = \frac{0 - 196}{20} = -9.8 \, \text{m/s}^2. \]With no tension, it is in free fall with acceleration equal to gravity downward.(d) For \( T = 196 \, \text{N} \):\[ a = \frac{196 - 196}{20} = 0. \]The crate is in equilibrium with no net force or acceleration.

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Key Concepts

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

Forces on Objects
Understanding the forces acting on an object is critical when predicting its motion. In this scenario, the forces involved include:
  • Gravity: A force pulling the object towards the Earth. For a 20-kg crate, this force is calculated by multiplying its mass by the acceleration due to gravity, which is approximately \(9.8 \, \text{m/s}^2\). Therefore, the gravitational force \(F_g\) acting downward is \(20 \, \text{kg} \times 9.8 \, \text{m/s}^2 = 196 \, \text{N}\).
  • Tension: This force acts along the rope, and its effect can be either upward, neutral, or non-existent if absent.
These forces are part of a fundamental study of physics and are key to understanding motion and acceleration of objects.
Tension in Physics
Tension is a force that arises in a rope or a cable when used to transmit pulling force. It plays a significant role in determining how an object moves. In our example of the hanging crate, the tension can vary, affecting how the crate accelerates:
  • When tension is greater than the gravitational force, the object moves upward.
  • If tension equals the gravitational force, the object remains stationary, in equilibrium.
  • When tension is lower than the gravitational pull, the object accelerates downward.
The calculation of tension involves examining both the direction and magnitude it contributes to the balance of forces applied to the object.
Acceleration Calculation
Acceleration measures how quickly the velocity of an object changes. It is determined by calculating the net force acting on the object and is described using Newton's Second Law.
Newton's Second Law states:\[F_{net} = m \cdot a\] In the case of the crate, the net force \(F_{net}\) is given by the tension in the rope minus the gravitational force:\[F_{net} = T - 196\]From this, acceleration \(a\) is calculated using:\[a = \frac{T - 196}{20}\] For different values of tension, the acceleration changes, illustrating how tension directly influences the object's movement.
Free Fall
Free fall is a condition where the only force acting on an object is gravity, making it an important concept in physics. When an object like our crate undergoes free fall, there is no tension in the supporting rope. This causes the object to accelerate downward with an acceleration equal to that of gravity, which is \(9.8 \, \text{m/s}^2\).
When \(T = 0 \, \text{N}\):
  • The net force is \(-196 \, \text{N}\) (gravity acting alone).
  • The acceleration \(a\) becomes \(-9.8 \, \text{m/s}^2\).
This situation exemplifies Newton's insight into how objects behave under the influence of just gravity, showing a pure gravitational acceleration unhindered by other forces.

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Most popular questions from this chapter

Find the weight on the surface of the Earth of a body whose mass is (a) \(3.00 \mathrm{~kg}\), and (b) \(200 \mathrm{~g}\). The general relation between mass \(m\) and weight \(F_{W}\) is \(F_{W}=m g\). In this relation, \(m\) must be in kilograms, \(g\) in meters per second squared, and \(F_{W}\) in newtons. On Earth, \(g=9.81 \mathrm{~m} / \mathrm{s} 2\). The acceleration due to gravity varies from place to place in the universe. (a) \(F_{W}=(3.00 \mathrm{~kg})\left(9.81 \mathrm{~m} / \mathrm{s}^{2}\right)=29.4 \mathrm{~kg} \cdot \mathrm{m} / \mathrm{s}^{2}=29.4 \mathrm{~N}\) (b) \(F_{W}=(0.200 \mathrm{~kg})\left(9.81 \mathrm{~m} / \mathrm{s}^{2}\right)=1.96 \mathrm{~N}\)

An 8000 -kg engine pulls a 40000 -kg train along a level track and gives it an acceleration \(a_{1}=1.20 \mathrm{~m} / \mathrm{s}^{2} .\) What acceleration \(\left(a_{2}\right)\) would the engine give to a 16000 -kg train? Ignore friction. For a given engine force, the acceleration is inversely proportional to the total mass. Thus, $$ a_{2}=\frac{m_{1}}{m_{2}} a_{1}=\frac{8000 \mathrm{~kg}+40000 \mathrm{~kg}}{8000 \mathrm{~kg}+16000 \mathrm{~kg}}\left(1.20 \mathrm{~m} / \mathrm{s}^{2}\right)=2.40 \mathrm{~m} / \mathrm{s}^{2} $$

Compute algebraically the resultant of the following coplanar forces: \(100 \mathrm{~N}\) at \(30^{\circ}, 141.4 \mathrm{~N}\) at \(45^{\circ}\), and \(100 \mathrm{~N}\) at \(240^{\circ}\). Check your result graphically.

A 600 -kg car is coasting along a level road at \(30 \mathrm{~m} / \mathrm{s}\). (a) How large a retarding force (assumed constant) is required to stop it in a distance of \(70 \mathrm{~m} ?(b)\) What is the minimum coefficient of friction between tires and roadway if this is to be possible? Assume the wheels are not locked, in which case we are dealing with static friction- there's no sliding. ( \(a\) ) First find the car's acceleration from a constant- \(a\) equation. It is known that \(\mathrm{v}_{i x}=30 \mathrm{v}_{f x},=0\), and \(x=70 \mathrm{~m}\). Use \(v_{k}^{2}=v_{u}^{2}+2 a x\) to find $$ a=\frac{v_{f x}^{2}-v_{i r}^{2}}{2 x}=\frac{0-900 \mathrm{~m}^{2} / \mathrm{s}^{2}}{140 \mathrm{~m}}=-6.43 \mathrm{~m} / \mathrm{s}^{2} $$ Now write $$ F=m a=(600 \mathrm{~kg})\left(-6.43 \mathrm{~m} / \mathrm{s}^{2}\right)-3858 \mathrm{~N} \text { or }-3.9 \mathrm{kN} $$ (b) Assume the force found in \((a)\) is supplied as the friction force between the tires and roadway. Therefore, the magnitude of the friction force on the tires is \(F_{\mathrm{f}}=3858 \mathrm{~N}\). The coefficient of friction is given by \(u=F_{f} / F_{N}\), where \(F_{N}\) is the normal force. In the present case, the roadway pushes up on the car with a force equal to the car's weight. Therefore, that $$ \begin{array}{l} F_{N}=F_{W}=m g=(600 \mathrm{~kg})\left(9.81 \mathrm{~m} / \mathrm{s}^{2}\right)=5886 \mathrm{~N} \\ \mu_{s}=\frac{F_{\mathrm{f}}}{F_{N}}=\frac{3858}{5886}=0.66 \end{array} $$ The coefficient of friction must be at least \(0.66\) if the car is to stop within \(70 \mathrm{~m}\).

The only force acting on a \(5.0\) -kg object has components \(F_{x}=20 \mathrm{~N}\) and \(F_{y}=30 \mathrm{~N}\). Find the acceleration of the object. Use \(\sum F_{x}=m a_{x}\) and \(\sum F_{y}=m a_{y}\) to obtain $$ \begin{array}{l} a_{x}=\frac{\sum F_{x}}{m}=\frac{20 \mathrm{~N}}{5.0 \mathrm{~kg}}=4.0 \mathrm{~m} / \mathrm{s}^{2} \\ a_{y}=\frac{\sum F_{y}}{m}=\frac{30 \mathrm{~N}}{5.0 \mathrm{~kg}}=6.0 \mathrm{~m} / \mathrm{s}^{2} \end{array} $$ These components of the acceleration are shown in Fig. \(3-10 .\) From the figure, $$ a=\sqrt{(4.0)^{2}+(6.0)^{2}} \mathrm{~m} / \mathrm{s}^{2}=7.2 \mathrm{~m} / \mathrm{s}^{2} $$ and \(\theta=\arctan (6.0 / 4.0)=56^{\circ}\).
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