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Inclined Plane Question. What force is necessary to keep the box from sliding?

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DavidAp
#1
Oct5-11, 06:16 AM
P: 44

"A loaded penguin sled weighing 69 N rests on a plane inclined at angle θ = 22 to the horizontal. Between the sled and the plane, the coefficient of static friction is 0.25, and the coefficient of kinetic friction is 0.20.

(a) What is the minimum magnitude of the force , 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) Whatvalue of F is required to move the sled up the plane at constant velocity?"

Answers:
(a) 9.85 N
(b) 41.8 N
(c) 38.6 N




Related Equations:
Mk = Ff/N
Ms = Fb/N


I started this problem by finding the normal and parallel force from the inclined plane.
N = 69cos(22) = 63.97N
Fp(arallel) = 69sin(22) = 25.85N


I then checked to see if the force pulling it down, parellel to the inclined plane, was greater or equal to the "budging" force to see if the box was even moving.

Fb = Fn(Ms) = 63.97N(0.25) = 15.99N

Seeing that 25.85N > 15.99N I proceeded to find the kinetic frictional force opposing the parallel force.

Ff = Fn(Mk) = 63.97N(0.2) = 12.97N

Therefore, the net force should be the parallel force pulling the box down the incline minus the kinetic frictional force opposing it.

Fnet = 25.85N - 12.79N = 13.05N

Now, in order to keep the box from sliding down the incline the amount of force I need to push on the box should be equal to the force causing the box to slide down, 13.05. However, the answer is 9.85N... and I'm way off!

What did I do wrong? Obviously something horrible went wrong in either my calculations or in my idea of how to get the answer... probably the latter. Can somebody explain to me what to do? Why what i did here didn't work?

Thank you for taking the time to review my question out of the sea of questions out there, I very much appreciate it.
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PeterO
#2
Oct5-11, 06:33 AM
HW Helper
P: 2,316
Quote Quote by DavidAp View Post

"A loaded penguin sled weighing 69 N rests on a plane inclined at angle θ = 22 to the horizontal. Between the sled and the plane, the coefficient of static friction is 0.25, and the coefficient of kinetic friction is 0.20.

(a) What is the minimum magnitude of the force , 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) Whatvalue of F is required to move the sled up the plane at constant velocity?"

Answers:
(a) 9.85 N
(b) 41.8 N
(c) 38.6 N




Related Equations:
Mk = Ff/N
Ms = Fb/N


I started this problem by finding the normal and parallel force from the inclined plane.
N = 69cos(22) = 63.97N
Fp(arallel) = 69sin(22) = 25.85N


I then checked to see if the force pulling it down, parellel to the inclined plane, was greater or equal to the "budging" force to see if the box was even moving.

Fb = Fn(Ms) = 63.97N(0.25) = 15.99N

Seeing that 25.85N > 15.99N I proceeded to find the kinetic frictional force opposing the parallel force.

Ff = Fn(Mk) = 63.97N(0.2) = 12.97N

Therefore, the net force should be the parallel force pulling the box down the incline minus the kinetic frictional force opposing it.

Fnet = 25.85N - 12.79N = 13.05N

Now, in order to keep the box from sliding down the incline the amount of force I need to push on the box should be equal to the force causing the box to slide down, 13.05. However, the answer is 9.85N... and I'm way off!

What did I do wrong? Obviously something horrible went wrong in either my calculations or in my idea of how to get the answer... probably the latter. Can somebody explain to me what to do? Why what i did here didn't work?

Thank you for taking the time to review my question out of the sea of questions out there, I very much appreciate it.
(a) Because you are preventing the sled from moving, you only have to support the static friction.
DavidAp
#3
Oct5-11, 06:47 AM
P: 44
Thank you so much!


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