# Friction of a sliding box on a slope

• J-dizzal
In summary, the problem involves a loaded penguin sled on an inclined plane with coefficients of static and kinetic friction. It asks for the minimum force required to prevent the sled from slipping down the plane, to start the sled moving up the plane, and to keep the sled moving up the plane at a constant velocity. The correct answers are 4.646 N, 14.276 N, and 7.663 N for parts a, b, and c respectively.

## Homework Statement

A loaded penguin sled weighing 75 N rests on a plane inclined at angle θ = 21° to the horizontal (see the figure). Between the sled and the plane, the coefficient of static friction is 0.30, and the coefficient of kinetic friction is 0.18. (a) What is the minimum magnitude of the force F , parallel to the plane, that will prevent the sled from slipping down the plane? (b) What is the minimum magnitude F that will start the sled moving up the plane? (c) What value of F is required to move the sled up the plane at constant velocity?

http://edugen.wileyplus.com/edugen/courses/crs7165/art/qb/qu/c06/fig06_29.gif

## Homework Equations

fss Fn is the magnitude of the maximum static force.
fkk Fn is the kinetic friction force.

## The Attempt at a Solution

http://s1294.photobucket.com/user/jhoversten/media/20150626_193750_zpsvqv6akeu.jpg.html [Broken]

i can see why my attempt at solving a is not working. thanks[/B]

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You did it right, except you miscalculated FT and FN

The angle is 21 degrees not 20 degrees.

opps thanks

Nathanael said:
You did it right, except you miscalculated FT and FN

The angle is 21 degrees not 20 degrees.
im getting 4.646 N but this is still wrong.

J-dizzal said:
im getting 4.646 N but this is still wrong.
That's not what I get. Did you correct both FT and FN?

Nathanael said:
That's not what I get. Did you correct both FT and FN?
oh yea i forgot to correct the right side of my equation. thanks

for part b; i set up and equation identical to the equation in part a except is used μk and got an answer 14.276 N. And this would be the force required to start the sled moving up the ramp after the static force was applied. I am not sure how to set it up

i added the force from part a to part b and that was not good.

Nathanael said:
That's not what I get. Did you correct both FT and FN?
i initiall thought it would be the same answer as part a, because once the block is broken free from gravitational forces and static frictional force it should be ready to move with any increase in force applied right?

J-dizzal said:
for part b; i set up and equation identical to the equation in part a except is used μk
μk does not come into play yet. You want to find the force F which puts it on the verge of slipping upwards, so you will still want to use μs

J-dizzal said:
i initiall thought it would be the same answer as part a, because once the block is broken free from gravitational forces and static frictional force it should be ready to move with any increase in force applied right?
It's similar to part a but it's not the same:
Friction opposes the motion of the block; in part a, the block was about slip downwards. In part b, the block is about to slip upwards.

Nathanael said:
μk does not come into play yet. You want to find the force F which puts it on the verge of slipping upwards, so you will still want to use μs

It's similar to part a but it's not the same:
Friction opposes the motion of the block; in part a, the block was about slip downwards. In part b, the block is about to slip upwards.
oh right friction is a non conservative force so it works in both directions.

J-dizzal said:
oh right friction is a non conservative force so it works in both directions.
Thanks Nathanael

No problem