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Chapter 3: Problem 19

A worker develops a tension \(T\) in the cable as he attempts to move the 50 -kg cart up the \(20^{\circ}\) incline. Determine the resulting acceleration of the cart if \((a) T=150 \mathrm{N}\) and \((b) T=200 \quad\) N. Neglect all friction, except that at the worker's feet.

Short Answer

Expert verified
(a) Acceleration is approximately 0.54 m/s虏, (b) Acceleration is approximately 1.54 m/s虏.

Step by step solution

01

Understand Forces Acting on the Cart

To determine the acceleration of the cart, we need to analyze the forces. The tension force, the gravitational force, and the normal force are acting on the cart. The tension force acts parallel to the incline, while the gravitational force acts downward, and it's component along the incline needs to be considered.
02

Calculate Gravitational Force Component

The gravitational force acting on the cart is given by \( F_g = mg \), where \( m = 50 \) kg and \( g = 9.8 \, \text{m/s}^2 \). The component of this force along the incline is \( F_{g, \text{parallel}} = mg \sin\theta \), where \( \theta = 20^{\circ} \).
03

Determine Net Force Along the Incline

The net force along the incline can be calculated using \( F_{\text{net}} = T - F_{g, \text{parallel}} \). For each given tension \( T \), calculate this net force.
04

Calculate Acceleration Using Newton's Second Law

Use Newton's Second Law, \( F = ma \), to find acceleration \( a \). Rearrange the equation to \( a = \frac{F_{\text{net}}}{m} \) using the net force calculated in the previous step.
05

Plug Values for Case (a) - Tension 150 N

First, calculate \( F_{g, \text{parallel}} = 50 \cdot 9.8 \cdot \sin(20^{\circ}) \). Then, find the net force \( F_{\text{net}} = 150 - F_{g, \text{parallel}} \) and use \( a = \frac{F_{\text{net}}}{50} \) to find the acceleration.
06

Plug Values for Case (b) - Tension 200 N

Repeat the process for \( T = 200 \) N. Calculate the net force \( F_{\text{net}} = 200 - F_{g, \text{parallel}} \), then \( a = \frac{F_{\text{net}}}{50} \) to find the acceleration.

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Key Concepts

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

Gravitational Force
Gravitational force is a key player in many physics problems. It is the force that the Earth exerts on any object with mass. To calculate this force, we use the formula \( F_g = mg \), where \( m \) is the mass and \( g \) is the acceleration due to gravity, roughly \( 9.8 \, \text{m/s}^2 \).
In problems involving inclined planes, it is crucial to determine the component of the gravitational force that acts along the plane. This component is calculated by projecting the gravitational force onto the plane, which is given by the formula \( F_{g, \text{parallel}} = mg \sin \theta \), where \( \theta \) is the angle of the incline.
By understanding how to decompose gravitational force into its components, you can tackle a variety of problems that involve slopes, ramps, or hills.
  • Gravitational force always acts downward towards the center of the Earth.
  • On an inclined plane, only the component parallel to the incline contributes to the motion along the plane.
Inclined Plane
An inclined plane is simply a flat surface tilted at an angle to the horizontal. Inclines are often used to study motion and forces because they introduce an interesting twist to typical linear motion.
The angle of the incline \( \theta \) plays a significant role. It affects the component of the gravitational force that acts parallel to the plane. The steeper the angle, the larger this component will be, and thus, it has a greater effect on movement down the incline.
  • Inclined planes help in studying forces like friction and tension in a controlled environment.
  • The angle of the plane directly influences the dynamics of the system.
Understanding these dynamics helps solve problems such as those involving ramps, slides, or any scenario where an object moves along a steep surface.
Tension Force
Tension force is a pulling force developed in cables, ropes, or strings when they bear load. In the given problem, the tension force helps the worker move the cart up the inclined plane.
The tension force acts parallel to the incline, opposing the gravitational pull along the plane. Therefore, analyzing its impact is necessary to determine the net force acting on the cart, which then allows you to apply Newton's Second Law to find acceleration.
The equations governing the scenario feature tension as a determining factor in finding out whether the net force will accelerate or decelerate the cart along the incline.
  • Tension opposes gravitational force when ascending an incline.
  • It is crucial for calculating net force and subsequent acceleration.
By understanding tension, we unlock solutions to many mechanical problems dealing with pulleys, hoists, and other lifting systems.

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Most popular questions from this chapter

It is experimentally determined that the drive wheels of a car must exert a tractive force of \(560 \mathrm{N}\) on the road surface in order to maintain a steady vehicle speed of \(90 \mathrm{km} / \mathrm{h}\) on a horizontal road. If it is known that the overall drivetrain efficiency is \(e_{m}=0.70,\) determine the required motor power output \(P\).

The 200 -kg glider \(B\) is being towed by airplane \(A\) which is flying horizontally with a constant speed of \(220 \mathrm{km} / \mathrm{h}\). The tow cable has a length \(r=60 \mathrm{m}\) and may be assumed to form a straight line. The glider is gaining altitude and when \(\theta\) reaches \(15^{\circ}\) the angle is increasing at the constant rate \(\dot{\theta}=\) 5 deg/s. At the same time the tension in the tow cable is \(1520 \mathrm{N}\) for this position. Calculate the aerodynamic lift \(L\) and \(\operatorname{drag} D\) acting on the glider.

Car \(A\) weighing 3200 lb and traveling north at \(20 \mathrm{mi} / \mathrm{hr}\) collides with car \(B\) weighing \(3600 \mathrm{lb}\) and traveling at \(30 \mathrm{mi} / \mathrm{hr}\) as shown. If the two cars become entangled and move together as a unit after the crash, compute the magnitude \(v\) of their common velocity immediately after the impact and the angle \(\theta\) made by the velocity vector with the north direction.

The position vector of a particle is given by \(\mathbf{r}=\) \(8 t \mathbf{i}+1.2 t^{2} \mathbf{j}-0.5\left(t^{3}-1\right) \mathbf{k},\) where \(t\) is the time in seconds from the start of the motion and where \(\mathbf{r}\) is expressed in meters. For the condition when \(t=\) 4 s, determine the power \(P\) developed by the force \(\mathbf{F}=40 \mathbf{i}-20 \mathbf{j}-36 \mathbf{k} \mathrm{N}\) which acts on the particle.

A boy weighing 100 lb runs and jumps on his \(20-1 b\) sled with a horizontal velocity of 15 ft/sec. If the sled and boy coast \(80 \mathrm{ft}\) on the level snow before coming to rest, compute the coefficient of kinetic friction \(\mu_{k}\) between the snow and the runners of the sled.
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