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Chapter 3: Problem 11

A \(5.0\) -kg object is to be given an upward acceleration of \(0.30 \mathrm{~m} / \mathrm{s}^{2}\) by a rope pulling straight upward on it. What must be the tension in the rope? The free-body diagram for the object is shown in Fig. \(3-8(b)\). The tension in the rope is \(F_{T}\), and the weight of the object is \(F_{W}=m g=\) \((5.0 \mathrm{~kg})\left(9.81 \mathrm{~m} / \mathrm{s}^{2}\right)=49.1 \mathrm{~N}\). Using \(\Sigma F_{y}=m a_{y}\) with up taken as positive, from which \(F_{T}=50.6 \mathrm{~N}=51 \mathrm{~N}\). As a check, we notice that \(F_{T}\) is larger than \(F_{W}\), as it must be if the object is to accelerate upward.

Short Answer

Expert verified
The tension in the rope must be 51 N.

Step by step solution

01

Identify Forces Acting on the Object

The object has two main forces acting on it: the tension in the rope, \(F_T\), pulling upward, and the gravitational force (weight), \(F_W\), pulling downward. The weight \(F_W = mg\), where \(m\) is the mass and \(g\) is the acceleration due to gravity \(9.81 \ \text{m/s}^2\).
02

Calculate the Weight of the Object

Calculate the weight of the object using the formula \(F_W = mg\). For a mass \(m = 5.0 \ \text{kg}\), \(F_W = (5.0 \ \text{kg})(9.81 \ \text{m/s}^2) = 49.1 \ \text{N}\).
03

Set Up Net Force Equation

The net force equation in the vertical direction is given by \(\Sigma F_y = ma_y\), where \(a_y\) is the acceleration. Here, \(F_T - F_W = ma_y\), and \(a_y = 0.30 \ \text{m/s}^2\).
04

Solve for Tension \(F_T\) in the Rope

Rearrange the net force equation to solve for \(F_T\): \(F_T = F_W + ma_y\). Substitute known values: \(F_T = 49.1 \ \text{N} + (5.0 \ \text{kg})(0.30 \ \text{m/s}^2)\).
05

Calculate Tension

Perform the calculation: \(F_T = 49.1 \ \text{N} + 1.5 \ \text{N} = 50.6 \ \text{N}\). Round off to two significant figures to get \(F_T \approx 51 \ \text{N}\).
06

Verify the Result

Check that the tension force \(F_T\) is greater than the weight \(F_W\) which indicates that the force is sufficient to provide upward acceleration. Thus \(F_T = 51 \ \text{N}\) is confirmed as correct.

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Key Concepts

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

Newton's Second Law
Newton's Second Law of Motion forms the foundation of many physics problems, especially in mechanics. It states that the net force acting on an object is equal to the mass of the object times its acceleration. This can be expressed by the equation \( \Sigma F = ma \), where \( \Sigma F \) is the sum of all forces acting on the object, \( m \) is the mass, and \( a \) is the acceleration. This relationship enables us to predict how an object will move when subjected to various forces. Newton's Second Law is crucial for calculating unknown forces in physics problems. It allows us to find quantities like tension in a rope when given acceleration and mass. In our problem, it helps determine how much force the rope must exert to accelerate the object upwards.
Free-Body Diagram
A free-body diagram is a powerful tool used in physics to visualize and analyze the forces acting on an object. It represents the object of interest as a dot or a simple box, and uses arrows to show the direction and magnitude of all forces applied.
  • These diagrams help in setting up the equations of motion like the one we used in this problem.
  • The length of each arrow reflects the force's magnitude.
  • Forces acting upwards or in positive coordinate directions are usually drawn above or to the right, while downward forces are below or to the left.
In the exercise, the free-body diagram should include an upward arrow representing the tension \( F_T \) and a downward arrow depicting the gravitational force \( F_W \). The simplicity of free-body diagrams makes them essential for understanding forces and predicting motion.
Tension in Ropes
The concept of tension in ropes is central to solving problems involving suspended objects. Tension is the force transmitted through a rope, string, or wire when it is pulled tight by forces acting from opposite ends. When calculating tension:
  • Remember that tension is uniform throughout the rope if it is massless and the only force acting in the system.
  • The tension needs to overcome other forces such as weight to lift the object upwards.
In our problem, the rope must exert an additional force beyond the object's weight to provide upward acceleration. This reveals why the calculated tension, 51 N, is greater than the weight, 49.1 N. Calculating tension accurately is vital for ensuring safety and efficiency in applications like engineering and construction.
Upward Acceleration
Upward acceleration occurs when an upward force exceeds the gravitational force acting on an object. In physics, we often take upward as the positive direction. So, when a net positive force is applied, the object accelerates upward. The acceleration dictates how quickly the object's velocity changes as it moves upward.
  • In our problem, the required upward acceleration is given as 0.30 m/s虏.
  • To achieve this acceleration, the net force must equip the object to overcome its weight and impart additional force.
The calculation for tension incorporates this additional needed force to facilitate upward movement. Understanding how forces contribute to acceleration is fundamental in designing systems that involve lifting or moving objects against gravity.

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Most popular questions from this chapter

Two forces, \(80 \mathrm{~N}\) and \(100 \mathrm{~N}\), acting at an angle of \(60^{\circ}\) with each other, pull on an object. (a) What single force would replace the two forces? (b) What single force (called the equilibrant) would balance the two forces? Solve algebraically.

A constant force acts on a \(5.0\) kg object and reduces its velocity from \(7.0 \mathrm{~m} / \mathrm{s}\) to \(3.0 \mathrm{~m} / \mathrm{s}\) in a time of \(3.0\) s. Determine the force. We must first find the acceleration of the object, which is constant because the force is constant. Taking the direction of motion as positive, from Chapter 2 $$ a=\frac{v_{f}-v_{i}}{t}=\frac{-4.0 \mathrm{~m} / \mathrm{s}}{3.0 \mathrm{~s}}=-1.33 \mathrm{~m} / \mathrm{s}^{2} $$ Use \(F=m a\) with \(m=5.0 \mathrm{~kg}\) : $$ F=(5.0 \mathrm{~kg})\left(-1.33 \mathrm{~m} / \mathrm{s}^{2}\right)=-6.7 \mathrm{~N} $$ The minus sign indicates that the force is a retarding force, directed opposite to the motion.

The Moon, whose mass is \(7.35 \times 10^{22} \mathrm{~kg}\), orbits the Earth, whose mass is \(5.98 \times 10^{24} \mathrm{~kg}\), at a mean distance of \(3.85 \times 10^{8} \mathrm{~m}\). It is held in a nearly circular orbit by the Earth-Moon gravitational interaction. Determine the force of gravity due to the planet acting on the Moon. From the universal law of gravitation $$ F_{G}=G \frac{m M}{R^{2}} $$ we get $$ F_{G}=6.673 \times 10^{-11} \mathrm{~N} \cdot \mathrm{m}^{2} / \mathrm{kg} \frac{\left(7.35 \times 10^{22} \mathrm{~kg}\right)\left(5.98 \times 10^{24}\right)}{\left(3.85 \times 10^{8} \mathrm{~m}\right)^{2}} $$ which yields $$ F_{G}=1.98 \times 10^{20} \mathrm{~N} $$ This is also the force on the Earth due to the Moon, and the force on the Moon due to the Earth.

A \(400-g\) block with an initial speed of \(80 \mathrm{~cm} / \mathrm{s}\) slides along a horizontal tabletop against a friction force of \(0.70 \mathrm{~N}\). \((a)\) How far will it slide before stopping? (b) What is the coefficient of friction between the block and the tabletop? (a) Take the direction of motion as positive. The only unbalanced force acting on the block is the friction force, \(-0.70 \mathrm{~N}\). Therefore, \(\Sigma F=m a \quad\) becomes \(\quad-0.70 \mathrm{~N}=(0.400 \mathrm{~kg})(a)\) from which \(a=-1.75 \mathrm{~m} / \mathrm{s}^{2}\). (Notice that \(m\) is always in kilograms.) To find the distance the block slides, make use of \(_{i x}\) \(=0.80 \mathrm{~m} / \mathrm{s}, f_{x}=0\), and \(a=-1.75 \mathrm{~m} / \mathrm{s}^{2} .\) Then \(v_{\mathrm{K}}^{2}-u_{u}^{2}=2 a x\) yields $$ x=\frac{v_{f x}^{2}-v_{i x}^{2}}{2 a}=\frac{(0-0.64) \mathrm{m}^{2} / \mathrm{s}^{2}}{(2)\left(-1.75 \mathrm{~m} / \mathrm{s}^{2}\right)}=0.18 \mathrm{~m} $$ (b) Because the vertical forces on the block must cancel, the upward push of the table \(F_{N}\) must equal the weight \(m g\) of the block. Then $$ \mu_{k}=\frac{\text { Friction force }}{F_{N}}=\frac{0.70 \mathrm{~N}}{(0.40 \mathrm{~kg})\left(9.81 \mathrm{~m} / \mathrm{s}^{2}\right)}=0.178 \text { or } 0.18 $$

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) \(196 \mathrm{~N}\).
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