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Chapter 5: Problem 96

Two blocks, with masses \(4.00 \mathrm{~kg}\) and \(8.00 \mathrm{~kg},\) are connected by a string and slide down a \(30.0^{\circ}\) inclined plane (Fig. \(\mathbf{P 5 . 9 6}\) ). The coefficient of kinetic friction between the \(4.00 \mathrm{~kg}\) block and the plane is \(0.25 ;\) that between the \(8.00 \mathrm{~kg}\) block and the plane is \(0.35 .\) Calculate (a) the acceleration of each block and (b) the tension in the string. (c) What happens if the positions of the blocks are reversed, so that the \(4.00 \mathrm{~kg}\) block is uphill from the \(8.00 \mathrm{~kg}\) block?

Short Answer

Expert verified
The acceleration of the block and the tension in the string can be calculated by considering the forces acting parallel to the incline and then utilizing Newton's second law. The entire setup would behave differently when the positions of the blocks are reversed due to change in static friction for the 4 kg block.

Step by step solution

01

Identify the Forces Involved

In this problem, there are two main forces acting on each block: the force of gravity (which can be divided into components perpendicular and parallel to the slope of the incline) and the force of friction which opposes the motion of the blocks. The force of gravity can be calculated as \( mg \), where \( m \) is the mass of the block and \( g \) is the acceleration due to gravity.
02

Calculate Combined Acceleration of Blocks

To calculate the combined acceleration of the blocks, we need to add the parallel forces and subtract the frictional forces.\nThe net force in the direction of motion can be calculated using Newton's second law: \( F=ma \). The forces acting along the incline include the gravitational force, \( mgsin(\Theta) \), and the static friction.\nSo, we have \n\( F_{net}=m_{total}a \)\n\( (m_{1}+m_{2}) a = m_{1} gsin(\Theta) - f_{1} + m_{2} gsin(\Theta) - f_{2} \)\nwhere \( m_{1}=4kg \), \( m_{2}=8kg \), \( f_{1}=0.25 \times m_{1}gcos(\Theta) \), \( f_{2}=0.35 \times m_{2}gcos(\Theta) \). Solving this we can find acceleration, \( a \).
03

Calculate Tension in the String

The tension in the string can be calculated by considering just one of the blocks (let's take the block with mass \( m_{2} = 8kg \)).\nUsing Newton's second law for \( m_{2} \) we get,\n\( T - m_{2} gsin(\Theta) + f_{2} = m_{2}a \)\nWe know the value of \( a \) from the previous step. Solving this equation, we can find the tension, \( T \).
04

Discuss Situation When Positions are Reversed

When positions of the blocks are reversed, the 4.00 kg block is uphill from the 8.00 kg block. This will mean that the smaller block will now have to overcome a greater static friction, since the larger block is now at the bottom. This might result in either a decrease in acceleration, or in the case of static friction being too high, no motion at all.

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Key Concepts

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

Newton's Second Law
Understanding how objects move and interact is at the core of physics, and Newton's Second Law of Motion is fundamental to this understanding. It states that the acceleration of an object is directly proportional to the net force acting upon it and inversely proportional to its mass. This can be summed up with the equation \( F = ma \), where \( F \) is the net force, \( m \) is the mass, and \( a \) is the acceleration.

When applying Newton's Second Law to the incline plane problem, we examine all forces affecting the motion of the blocks down the slope. These forces must be multiplied by the mass to determine the resulting acceleration. In this scenario, the net force and the resulting acceleration act along the incline, and all opposing forces, such as friction, subtract from the total force to influence the final acceleration.
Kinetic Friction
Friction is the resisting force encountered by an object when it moves over a surface. Kinetic friction, specifically, acts on objects that are already in motion. It is described by the formula \( f_k = \mu_k N \), where \( \mu_k \) is the coefficient of kinetic friction and \( N \) is the normal force鈥攖he perpendicular force the surface exerts on the object.

In the problem of blocks sliding on an incline, the kinetic friction works against the gravitational force pulling the blocks downhill. The frictional forces for each block are different due to varying coefficients of friction. This is important to include in the total force calculation because kinetic friction reduces the blocks' acceleration down the slope, affecting both the tension in the string and the ultimate motion of the system.
Gravitational Force
The force of gravity is ever-present and affects every object with mass. On Earth, this force is what gives weight to objects and causes them to fall towards the ground. The gravitational force on an object is calculated with the formula \( F_g = mg \), where \( m \) is the mass and \( g \) is the acceleration due to gravity, approximately \( 9.8 m/s^2 \) at Earth's surface.

For inclined planes, gravity's effect is split into two components: one perpendicular to the plane, and one parallel to the plane. The component parallel to the plane \( (mg \sin(\Theta)) \) is what drives the blocks to slide downwards. This force, combined with kinetic friction and the mass of the blocks, is essential for calculating the blocks' acceleration and the tension in the string connecting them.
Tension in a String
Tension is a force that is transmitted through a string, cable, or rope when it is pulled tight by forces acting from either end. In the problem context, the tension in the string connecting the two blocks counteracts some of the gravitational force pulling the blocks down the slope. The formula to calculate tension is not straightforward, as it involves evaluating the sum of the forces acting on one of the blocks.

By focusing on just one block, we can isolate the tension in the string and apply Newton鈥檚 second law. The tension calculation must consider the gravitational pull and the kinetic friction for that particular block. Tension ensures that the two blocks accelerate together, and altering this tension will impact the overall dynamics of the system on the incline.

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Most popular questions from this chapter

DATA You are an engineer working for a manufacturing company. You are designing a mechanism that uses a cable to drag heavy metal blocks a distance of \(8.00 \mathrm{~m}\) along a ramp that is sloped at \(40.0^{\circ}\) above the horizontal. The coefficient of kinetic friction between these blocks and the incline is \(\mu_{\mathrm{k}}=0.350 .\) Each block has a mass of \(2170 \mathrm{~kg}\). The block will be placed on the bottom of the ramp, the cable will be attached, and the block will then be given just enough of a momentary push to overcome static friction. The block is then to accelerate at a constant rate to move the \(8.00 \mathrm{~m}\) in \(4.20 \mathrm{~s}\). The cable is made of wire rope and is parallel to the ramp surface. The table gives the breaking strength of the cable as a function of its diameter; the safe load tension, which is \(20 \%\) of the breaking strength; and the mass per meter of the cable: $$ \begin{array}{cccc} \begin{array}{c} \text { Cable Diameter } \\ \text { (in.) } \end{array} & \begin{array}{c} \text { Breaking Strength } \\ (\mathbf{k N}) \end{array} & \begin{array}{c} \text { Safe Load } \\ (\mathbf{k N}) \end{array} & \begin{array}{c} \text { Mass per Meter } \\ (\mathbf{k g} / \mathbf{m}) \end{array} \\ \hline \frac{1}{4} & 24.4 & 4.89 & 0.16 \\ \frac{3}{8} & 54.3 & 10.9 & 0.36 \\ \frac{1}{2} & 95.2 & 19.0 & 0.63 \\ \frac{5}{8} & 149 & 29.7 & 0.98 \\ \frac{3}{4} & 212 & 42.3 & 1.41 \\ \frac{7}{8} & 286 & 57.4 & 1.92 \\ 1 & 372 & 74.3 & 2.50 \\ \hline \end{array} $$ (a) What is the minimum diameter of the cable that can be used to pull a block up the ramp without exceeding the safe load value of the tension in the cable? Ignore the mass of the cable, and select the diameter from those listed in the table. (b) You need to know safe load values for diameters that aren't in the table, so you hypothesize that the breaking strength and safe load limit are proportional to the cross-sectional area of the cable. Draw a graph that tests this hypothesis, and discuss its accuracy. What is your estimate of the safe load value for a cable with diameter \(\frac{9}{16}\) in.? (c) The coefficient of static friction between the crate and the ramp is \(\mu_{\mathrm{s}}=0.620,\) which is nearly twice the value of the coefficient of kinetic friction. If the machinery jams and the block stops in the middle of the ramp, what is the tension in the cable? Is it larger or smaller than the value when the block is moving? (d) Is the actual tension in the cable, at its upper end, larger or smaller than the value calculated when you ignore the mass of the cable? If the cable is \(9.00 \mathrm{~m}\) long, how accurate is it to ignore the cable's mass?

Apparent Weight. A \(550 \mathrm{~N}\) physics student stands on a bathroom scale in an elevator that is supported by a cable. The combined mass of student plus elevator is \(850 \mathrm{~kg}\). As the elevator starts moving, the scale reads \(450 \mathrm{~N}\). (a) Find the acceleration of the elevator (magnitude and direction). (b) What is the acceleration if the scale reads \(670 \mathrm{~N}\) ? (c) If the scale reads zero, should the student worry? Explain. (d) What is the tension in the cable in parts (a) and (c)?

A box with mass \(m\) sits at the bottom of a long ramp that is sloped upward at an angle \(\alpha\) above the horizontal. You give the box a quick shove, and after it leaves your hands it is moving up the ramp with an initial speed \(v_{0}\). The box travels a distance \(d\) up the ramp and then slides back down. When it returns to its starting point, the speed of the box is half the speed it started with; it has speed \(v_{0} / 2 .\) What is the coefficient of kinetic friction between the box and the ramp? (Your answer should depend on only \(\alpha\).)

A \(2540 \mathrm{~kg}\) test rocket is launched vertically from the launch pad. Its fuel (of negligible mass) provides a thrust force such that its vertical velocity as a function of time is given by \(v(t)=A t+B t^{2}\), where \(A\) and \(B\) are constants and time is measured from the instant the fuel is ignited. The rocket has an upward acceleration of \(1.50 \mathrm{~m} / \mathrm{s}^{2}\) at the instant of ignition and, 1.00 s later, an upward velocity of \(2.00 \mathrm{~m} / \mathrm{s}\). (a) Determine \(A\) and \(B\), including their SI units. (b) At 4.00 s after fuel ignition, what is the acceleration of the rocket, and (c) what thrust force does the burning fuel exert on it, assuming no air resistance? Express the thrust in newtons and as a multiple of the rocket's weight. (d) What was the initial thrust due to the fuel?

\(\mathrm{CP}\) A \(5.00 \mathrm{~kg}\) box sits at rest at the bottom of a ramp that is \(8.00 \mathrm{~m}\) long and is inclined at \(30.0^{\circ}\) above the horizontal. The coefficient of kinetic friction is \(\mu_{\mathrm{k}}=0.40,\) and the coefficient of static friction is \(\mu_{\mathrm{s}}=0.43 .\) What constant force \(F,\) applied parallel to the surface of the ramp, is required to push the box to the top of the ramp in a time of \(6.00 \mathrm{~s} ?\)
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