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Chapter 4: Problem 19

A mass slides on a ramp that is at an angle of \(\theta\) above the horizontal. The coefficient of friction between the mass and the ramp is \(\mu\). a) Find an expression for the magnitude and direction of the acceleration of the mass as it slides up the ramp. b) Repeat part (a) to find an expression for the magnitude and direction of the acceleration of the mass as it slides down the ramp.

Short Answer

Expert verified
a) When the mass slides up the ramp, the magnitude of the acceleration is \(a_{up} = g \sin{\theta} - \mu g \cos{\theta}\), and its direction is opposite to the direction of motion. b) When the mass slides down the ramp, the magnitude of the acceleration is \(a_{down} = g \sin{\theta} + \mu g \cos{\theta}\), and its direction is along the direction of motion.

Step by step solution

01

Find the components of the gravitational force

The gravitational force applied to the mass can be divided into two components along the parallel and perpendicular axes of the ramp. \(F_{g \parallel} = m g \sin{\theta}\) \(F_{g \perp} = m g \cos{\theta}\), where m is the mass of the object, g is the acceleration due to gravity, and \(\theta\) is the angle of the ramp above the horizontal.
02

Determine the force of friction

The force of friction (\(F_f\)) between the mass and the ramp is given by the product of the coefficient of friction (\(\mu\)) and the normal force (\(N\)) exerted on the mass. In this case, the normal force is equal to the perpendicular component of the gravitational force (\(F_{g \perp}\)). \(F_f = \mu N = \mu F_{g \perp} = \mu m g \cos{\theta}\)
03

Calculate the net force for the mass sliding up the ramp

When the mass slides up the ramp, the force of friction acts in the opposite direction to the parallel component of the gravitational force, so we subtract the friction force from the gravitational force parallel. \(F_{net \, up} = F_{g \parallel} - F_f = m g \sin{\theta} - \mu m g \cos{\theta}\)
04

Calculate the acceleration for the mass sliding up the ramp

To find the acceleration of the mass sliding up the ramp, divide the net force for the mass sliding up the ramp by the mass (m): \(a_{up} = \frac{F_{net \, up}}{m} = g \sin{\theta} - \mu g \cos{\theta}\) The direction of the acceleration is opposite to the direction of motion (up the ramp), as the mass is slowing down.
05

Calculate the net force for the mass sliding down the ramp

When the mass slides down the ramp, the force of friction acts in the opposite direction to the motion, which is also the same direction as the gravitational force acting parallel to the ramp. So, we need to subtract the friction force from the gravitational force parallel: \(F_{net \, down} = F_{g \parallel} + F_f = m g \sin{\theta} + \mu m g \cos{\theta}\)
06

Calculate the acceleration for the mass sliding down the ramp

To find the acceleration of the mass sliding down the ramp, divide the net force for the mass sliding down the ramp by the mass (m): \(a_{down} = \frac{F_{net \, down}}{m} = g \sin{\theta} + \mu g \cos{\theta}\) The direction of the acceleration is along the direction of motion (down the ramp), as the mass is speeding up. In conclusion: a) The magnitude of the acceleration of the mass as it slides up the ramp is \(a_{up} = g \sin{\theta} - \mu g \cos{\theta}\) and its direction is opposite to the direction of motion. b) The magnitude of the acceleration of the mass as it slides down the ramp is \(a_{down} = g \sin{\theta} + \mu g \cos{\theta}\) and its direction is along the direction of motion.

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Most popular questions from this chapter

4.32 A hanging mass, \(M_{1}=0.50 \mathrm{~kg}\), is attached by a light string that runs over a frictionless pulley to a mass \(M_{2}=1.50 \mathrm{~kg}\) that is initially at rest on a frictionless table. Find the magnitude of the acceleration, \(a,\) of \(M_{2}\)

Four weights, of masses \(m_{1}=6.50 \mathrm{~kg}_{3}\) \(m_{2}=3.80 \mathrm{~kg}, m_{3}=10.70 \mathrm{~kg},\) and \(m_{4}=\)

-4.45 A pinata of mass \(M=8.0 \mathrm{~kg}\) is attached to a rope of negligible mass that is strung between the tops of two vertical poles. The horizontal distance between the poles is \(D=2.0 \mathrm{~m},\) and the top of the right pole is a vertical distance \(h=0.50 \mathrm{~m}\) higher than the top of the left pole. The pinata is attached to the rope at a horizontal position halfway between the two poles and at a vertical distance \(s=1.0 \mathrm{~m}\) below the top of the left pole. Find the tension in each part of the rope due to the weight of the pinata.

Which of the following observations about the friction force is (are) incorrect? a) The magnitude of the kinetic friction force is always proportional to the normal force b) The magnitude of the static friction force is always proportional to the normal force. c) The magnitude of the static friction force is always proportional to the external applied force. d) The direction of the kinetic friction force is always opposite the direction of the relative motion of the object writh respect to the surface the object moves on. e) The direction of the static friction force is always opposite that of the impending motion of the object relative to the surface it rests on f) All of the above are correct.

Two blocks of masses \(m_{1}\) and \(m_{2}\) are suspended by a massless string over a frictionless pulley with negligible mass, as in an Atwood machine. The blocks are held motionless and then released. If \(m_{1}=3.50 \mathrm{~kg}\). what value does \(m_{2}\) have to have in order for the system to experience an acceleration \(a=0.400 g\) ? (Hint: There are two solutions to this problem.
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