Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 4: Problem 112

You want to push a \(65-\mathrm{kg}\) box up a \(25^{\circ}\) ramp. The coefficient of kinetic friction between the ramp and the box is \(0.30 .\) With what magnitude force parallel to the ramp should you push on the box so that it moves up the ramp at a constant speed?

Short Answer

Expert verified
Answer: To push the box up the ramp at a constant speed, you should apply a force of approximately 249.5 N parallel to the ramp.

Step by step solution

01

Identify the forces acting on the box

First, let's decompose the gravitational force acting on the box into two components: one parallel to the ramp and another perpendicular to it. The force of gravity is given by \(F_g = mg\), where m is the mass of the box, and g is the acceleration due to gravity. The component parallel to the ramp will be \(F_{g\parallel} = mg\sin(25^{\circ})\) and the component perpendicular to the ramp will be \(F_{g\perp} = mg\cos(25^{\circ})\). Also, note that the force of friction is given by \(F_f = \mu F_N\), where \(F_N\) is the normal force acting on the box, and \(\mu\) is the coefficient of kinetic friction.
02

Apply Newton's second law of motion

Since the box moves up the ramp at a constant speed, the net force along the ramp is zero. Therefore, the applied force (\(F_{app}\)) is equal to the sum of the component of the gravitational force parallel to the ramp and the frictional force: \(F_{app} = F_{g\parallel} + F_f\).
03

Calculate the normal force

The normal force acting on the box is equal to the component of the gravitational force perpendicular to the ramp: \(F_N = F_{g\perp} = mg\cos(25^{\circ})\).
04

Calculate the frictional force

Now that we know the normal force, we can find the force of friction: \(F_f = \mu F_N = \mu mg\cos(25^{\circ})\).
05

Calculate the applied force

Finally, plug the values for \(F_{g\parallel}\) and \(F_f\) into the equation for the applied force: \(F_{app} = mg\sin(25^{\circ}) + \mu mg\cos(25^{\circ})\). Using the given values for the mass (\(m\)), the angle of the ramp (\(25^{\circ}\)), and the coefficient of kinetic friction (\(\mu\)), we can compute the applied force: \(F_{app} = (65\text{ kg})(9.81\frac{\text{m}}{\text{s}^2})\sin(25^{\circ}) + 0.30 (65\text{ kg})(9.81\frac{\text{m}}{\text{s}^2})\cos(25^{\circ})\) Upon calculation, we find that \(F_{app} \approx 249.5\text{ N}\). Therefore, you should push the box with a force of approximately \(249.5\text{ N}\) parallel to the ramp in order for it to move up the ramp at a constant speed.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Start your free trial

Over 30 million students worldwide already upgrade their learning with Vaia!

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Before hanging new William Morris wallpaper in her bedroom, Brenda sanded the walls lightly to smooth out some irregularities on the surface. The sanding block weighs \(2.0 \mathrm{N}\) and Brenda pushes on it with a force of $3.0 \mathrm{N}\( at an angle of \)30.0^{\circ}$ with respect to the vertical, and angled toward the wall. Draw an FBD for the sanding block as it moves straight up the wall at a constant speed. What is the coefficient of kinetic friction between the wall and the block?
The mass of the Moon is 0.0123 times that of the Earth. A spaceship is traveling along a line connecting the centers of the Earth and the Moon. At what distance from the Earth does the spaceship find the gravitational pull of the Earth equal in magnitude to that of the Moon? Express your answer as a percentage of the distance between the centers of the two bodies.
The forces on a small airplane (mass \(1160 \mathrm{kg}\) ) in horizontal flight heading eastward are as follows: gravity \(=16.000 \mathrm{kN}\) downward, lift \(=16.000 \mathrm{kN}\) upward, thrust \(=1.800 \mathrm{kN}\) eastward, and \(\mathrm{drag}=1.400 \mathrm{kN}\) westward. At \(t=0,\) the plane's speed is \(60.0 \mathrm{m} / \mathrm{s} .\) If the forces remain constant, how far does the plane travel in the next \(60.0 \mathrm{s} ?\)
A bag of potatoes with weight \(39.2 \mathrm{N}\) is suspended from a string that exerts a force of \(46.8 \mathrm{N}\). If the bag's acceleration is upward at \(1.90 \mathrm{m} / \mathrm{s}^{2},\) what is the mass of the potatoes?
A skydiver, who weighs \(650 \mathrm{N},\) is falling at a constant speed with his parachute open. Consider the apparatus that connects the parachute to the skydiver to be part of the parachute. The parachute pulls upward with a force of \(620 \mathrm{N} .\) (a) What is the force of the air resistance acting on the skydiver? (b) Identify the forces and the interaction partners of each force exerted on the skydiver. (c) Identify the forces and interaction partners of each force exerted on the parachute.
See all solutions

Recommended explanations on Physics Textbooks

Physical Quantities And Units

Read Explanation

Torque and Rotational Motion

Read Explanation

Magnetism and Electromagnetic Induction

Read Explanation

Work Energy and Power

Read Explanation

Atoms and Radioactivity

Read Explanation

Measurements

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free