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Chapter 5: Problem 71

When a crate with mass \(25.0 \mathrm{~kg}\) is placed on a ramp that is inclined at an angle \(\alpha\) below the horizontal, it slides down the ramp with an acceleration of \(4.9 \mathrm{~m} / \mathrm{s}^{2} .\) The ramp is not friction less. To increase the acceleration of the crate, a downward vertical force \(\overrightarrow{\boldsymbol{F}}\) is applied to the top of the crate. What must \(F\) be in order to increase the acceleration of the crate so that it is \(9.8 \mathrm{~m} / \mathrm{s}^{2}\) ? How does the value of \(F\) that you calculate compare to the weight of the crate?

Short Answer

Expert verified
The additional force \(F\) needed to increase the acceleration of the crate to \(9.8 m/s^2\) is \(122.5 N\). This force is approximately half of the crate's weight (\(245 N\)).

Step by step solution

01

Identify the given variables

From the problem, it is given that the mass \(m = 25.0 \mathrm{~kg}\), acceleration \(a_1 = 4.9 \mathrm{~m/s^2}\), and the desired acceleration \(a_2 = 9.8 \mathrm{~m/s^2}\). The acceleration due to gravity \( g = 9.8 \mathrm{~m/s^2}\) is a known constant.
02

Determine the force due to gravity

The net force acting on the crate without any external force can be calculated using following equation, \(F_{net1} = m.a_1\). Let's calculate the net force. \( F_{net1} = 25 \times 4.9 = 122.5 N\). This \(F_{net1}\) is result of force due to gravity and force as a result of ramp inclination.
03

Calculate the force due to gravity

Force due to gravity is given as \( F_{gravity} = m \times g \). So, \(F_{gravity} = 25 \times 9.8 = 245 N\). Also from trigonometry we know that, force due to gravity acting down the slide is \(F_{gravity} \times sin(\alpha)\). Given, \(F_{gravity} \times sin(\alpha) = F_{net1}\) we can calculate angle \(\alpha = sin^{-1}( \frac{F_{net1}}{F_{gravity}} ) = sin^{-1}( \frac{122.5}{245} ) = 30\degree\)
04

Determine the additional force

Now, the crate should accelerate at \(a_2 = 9.8 \mathrm{~m/s^2}\), which requires a larger net force \(F_{net2}\) obeying the relation \(F_{net2} = m.a_2\). So, \( F_{net2} = 25 \times 9.8 = 245 N\). Hence, the additional force \(F\) to be added equals \(F_{net2} - F_{net1} = 245 - 122.5 = 122.5 N \). This calculated force should be added to the crate.
05

Compare F with the weight of the crate

The weight of the crate is equivalent to the force of gravity on it (\(F_{gravity} = 245 N\)). Hence, compared to the weight of the crate, the extra force \(F = 122.5 N\) applied is approximately half.

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Key Concepts

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

Inclined Plane
An inclined plane is a flat surface tilted at an angle to the horizontal. It is commonly used to move heavy objects upwards or downwards with less force than lifting the object directly. In our exercise, the crate is placed on an inclined plane that creates a unique interaction between gravity and movement.

One of the primary effects of using an inclined plane is that it splits the force of gravity into two components:
  • Parallel to the surface, which pushes the object down the slope.
  • Perpendicular to the surface, which presses the object into the plane.
The angle of inclination (\(\alpha\)) plays a key role in calculating these components. For our example, the angle was found to be \( 30^{\circ} \) using trigonometric relationships.

As objects slide down a ramp, friction and the angle of the ramp together determine the ease of movement. The optimal inclination allows movement without requiring much extra effort to overcome friction.
Friction and Forces
When analyzing an inclined plane with friction, understanding the interplay between frictional force and gravity is crucial. Friction opposes motion, arising from the contact between surfaces. On an inclined plane, this frictional force is typically a fraction of the normal force, which is the component of gravity acting perpendicular to the ramp.

In the case of the crate, the net force acting on it without external forces was determined by the equation\(F_{net1} = m \cdot a_1\), where \(a_1\) is the initial acceleration. This force is a combination of the force of gravity along the incline and frictional resistance. The friction here resisted the free slide of the crate, which is why an additional force was required to achieve a greater acceleration.

If there were no friction, the crate would accelerate only due to gravity, with no extra force needed. However, in most real-world scenarios, friction cannot be ignored, as it significantly impacts the force calculations for moving objects down inclines.
Acceleration Calculation
Calculating acceleration on an inclined plane involves Newton's Second Law of Motion, which links force, mass, and acceleration with the formula \(F = m \cdot a\). This exercise focused on increasing the crate's acceleration from \(4.9 \ m/s^2\) to \(9.8 \ m/s^2\). To achieve this, additional external force was necessary.

The initial force (without extra influence) allowed for \(F_{net1} = 122.5 \ N\) based on the given acceleration. To enhance the acceleration to \(9.8 \ m/s^2\), the net force needed to be \(F_{net2} = 245 \ N\). The additional applied force ensured that the crate could overcome the ramp's friction more effectively, bringing about the desired acceleration.

This calculated force was determined as \(122.5 \ N\). It illustrates the practical application of physics principles, demonstrating how calculated forces directly influence movement in physical environments.

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Most popular questions from this chapter

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 \(2.00 \mathrm{~kg}\) box is moving to the right with speed \(9.00 \mathrm{~m} / \mathrm{s}\) on a horizontal, friction less surface. At \(t=0\) a horizontal force is applied to the box. The force is directed to the left and has magnitude \(F(t)=\left(6.00 \mathrm{~N} / \mathrm{s}^{2}\right) t^{2} .\) (a) What distance does the box move from its position at \(t=0\) before its speed is reduced to zero? (b) If the force continues to be applied, what is the speed of the box at \(t=3.00 \mathrm{~s} ?\)

You are lowering two boxes, one on top of the other, down a ramp by pulling on a rope parallel to the surface of the ramp (Fig. E5.33). Both boxes move together at a constant speed of \(15.0 \mathrm{~cm} / \mathrm{s}\). The coefficient of kinetic friction between the ramp and the lower box is 0.444 , and the coefficient of static friction between the two boxes is 0.800 . (a) What force do you need to exert to accomplish this? (b) What are the magnitude and direction of the friction force on the upper box?

Two \(25.0 \mathrm{~N}\) weights are suspended at opposite ends of a rope that passes over a light, friction less pulley. The pulley is attached to a chain from the ceiling. (a) What is the tension in the rope? (b) What is the tension in the chain?

DATA In your physics lab, a block of mass m is at rest on a horizontal surface. You attach a light cord to the block and apply a horizontal force to the free end of the cord. You find that the block remains at rest until the tension \(T\) in the cord exceeds \(20.0 \mathrm{~N}\). For \(T>20.0 \mathrm{~N},\) you measure the acceleration of the block when \(T\) is maintained at a constant value, and you plot the results (Fig. \(\mathrm{P} 5.109)\). The equation for the straight line that best fits your data is \(a=\left[0.182 \mathrm{~m} /\left(\mathrm{N} \cdot \mathrm{s}^{2}\right)\right] T-2.842 \mathrm{~m} / \mathrm{s}^{2}\) For this block and surface, what are (a) the coefficient of static friction and (b) the coefficient of kinetic friction? (c) If the experiment were done on the earth's moon, where \(g\) is much smaller than on the earth, would the graph of \(a\) versus \(T\) still be fit well by a straight line? If so, how would the slope and intercept of the line differ from the values in Fig. \(\mathrm{P} 5.109 ?\) Or, would each of them be the same?
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