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Chapter 4: Problem 54

A crate full of artichokes rests on a ramp that is inclined \(10.0^{\circ}\) above the horizontal. Give the direction of the normal force and the friction force acting on the crate in each of these situations. (a) The crate is at rest. (b) The crate is being pushed and is sliding up the ramp. (c) The crate is being pushed and is sliding down the ramp.

Short Answer

Expert verified
Answer: The normal force always acts perpendicular to the ramp, making an angle of \(10.0^{\circ}\) with the horizontal. The friction force opposes the motion of the crate: when the crate is at rest, it acts upward along the ramp; when the crate is sliding up the ramp, it acts down the ramp; and when the crate is sliding down the ramp, it acts up the ramp.

Step by step solution

01

Situation (a) - The crate is at rest.

When the crate is at rest, the forces on it must be balanced in both the horizontal and vertical directions. The normal force acts perpendicular to the ramp and therefore makes an angle of \(10.0^{\circ}\) with the horizontal. The friction force opposes the potential motion of the crate (which would be downwards), so it acts in a direction opposite to that, which is upward along the ramp.
02

Situation (b) - The crate is being pushed and is sliding up the ramp.

When the crate is being pushed and sliding up the ramp, the normal force again acts perpendicular to the ramp, making an angle of \(10.0^{\circ}\) with the horizontal. However, for this scenario, the friction force acts in a direction opposite to the motion of the crate, which is down the ramp.
03

Situation (c) - The crate is being pushed and is sliding down the ramp.

When the crate is being pushed and sliding down the ramp, the normal force still acts perpendicular to the ramp, making an angle of \(10.0^{\circ}\) with the horizontal. The frictional force opposes the motion of the crate and acts in the direction opposite to that of the crate's motion, which is up the ramp. In summary, the direction of the normal force is always perpendicular to the ramp and makes an angle of \(10.0^{\circ}\) with the horizontal. The direction of the friction force depends on the motion of the crate and will act to oppose the direction of the crate's motion, either up or down the ramp.

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Most popular questions from this chapter

A crate of potatoes of mass \(18.0 \mathrm{kg}\) is on a ramp with angle of incline \(30^{\circ}\) to the horizontal. The coefficients of friction are \(\mu_{\mathrm{s}}=0.75\) and \(\mu_{\mathrm{k}}=0.40 .\) Find the frictional force (magnitude and direction) on the crate if the crate is sliding up the ramp.
A block of mass \(m_{1}=3.0 \mathrm{kg}\) rests on a frictionless horizontal surface. A second block of mass \(m_{2}=2.0 \mathrm{kg}\) hangs from an ideal cord of negligible mass that runs over an ideal pulley and then is connected to the first block. The blocks are released from rest. (a) Find the acceleration of the two blocks after they are released. (b) What is the velocity of the first block 1.2 s after the release of the blocks, assuming the first block does not run out of room on the table and the second block does not land on the floor? (c) How far has block 1 moved during the 1.2 -s interval? (d) What is the displacement of the blocks from their initial positions 0.40 s after they are released?
A truck driving on a level highway is acted on by the following forces: a downward gravitational force of \(52 \mathrm{kN}\) (kilonewtons); an upward contact force due to the road of \(52 \mathrm{kN} ;\) another contact force due to the road of \(7 \mathrm{kN},\) directed east; and a drag force due to air resistance of \(5 \mathrm{kN}\), directed west. What is the net force acting on the truck?
Two blocks are connected by a lightweight, flexible cord that passes over a frictionless pulley. If \(m_{1}=3.6 \mathrm{kg}\) and \(m_{2}=9.2 \mathrm{kg},\) and block 2 is initially at rest \(140 \mathrm{cm}\) above the floor, how long does it take block 2 to reach the floor?
A 6.0 -kg block, starting from rest, slides down a frictionless incline of length \(2.0 \mathrm{m} .\) When it arrives at the bottom of the incline, its speed is \(v_{\mathrm{f}} .\) At what distance from the top of the incline is the speed of the block \(0.50 v_{\mathrm{f}} ?\)
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