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Block on an accelerating ramp

by gaobo9109
Tags: accelerating, block, ramp
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gaobo9109
#1
Mar19-10, 08:42 PM
P: 66
1. The problem statement, all variables and given/known data
A block of mass m is initially at rest at the top of a ramp, inclined at angle θ to the horizontal. The coefficient of friction between the block and the ramp is Á. The ramp is now pushed at acceleration a to the right. For what range of value of a will the block start slipping down the ramp?


2. Relevant equations
f ≤ ÁN


3. The attempt at a solution
I think to tackle this question i need to find out the value of a for which the block will remain stationary. Help me check if my working is correct.

If the block doesn't slip, then its acceleration would also be a. The component of acceleration parallel to the plane would be acosθ, the component normal to the plane would be asinθ.

From here I formed the equation:
f - mgsinθ = macosθ
f ≤ mgcosθ + masinθ

I am not sure if this is the correct reasoning.
1. The problem statement, all variables and given/known data



2. Relevant equations



3. The attempt at a solution
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ehild
#2
Mar20-10, 01:06 AM
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Your method is correct in principle, but you have to include the normal force in the equation for the normal components.


ehild
gaobo9109
#3
Mar20-10, 01:56 AM
P: 66
From the answer, the range of value of a is from g(Ácosθ-sinθ)/(cosθ + Ásinθ) to gcotθ. So there must be two conditions for equilibrium. But from my two equations, i can only obtain one value of a. How do I find the other value of a for which the block is in equilibrium?

ehild
#4
Mar20-10, 02:31 AM
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Block on an accelerating ramp

If the block is in equilibrium, it can move neither up or down along the ramp. The direction of friction is opposite in these cases.

ehild
gaobo9109
#5
Mar20-10, 05:46 AM
P: 66
But i have a problem. It is the frictional force that move the block forward when the ramp is in motion. So how can frictional force be down the ramp, which is in the opposite direction of motion?
ehild
#6
Mar21-10, 01:24 AM
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Quote Quote by gaobo9109 View Post
But i have a problem. It is the frictional force that move the block forward when the ramp is in motion. So how can frictional force be down the ramp, which is in the opposite direction of motion?
It is both the normal force and friction that affect the horizontal motion of the block.

Don't you have a picture? The problem says that the ramp accelerates to the right, but I do not know if the ramp has its highest point on the right or on the left.

ehild
gaobo9109
#7
Mar22-10, 10:15 AM
P: 66
The highest point of the ramp is on the right. But i still don't get it. I thought normal force would not affect the acceleration parallel to the plane of incline. If normal force actually comes into play, what would the new equations for equilibrium be?
ehild
#8
Mar22-10, 10:38 AM
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The normal force does affect the acceleration normal to the incline. You need to set up two equations: one between the component of acceleration and forces parallel to the incline, and the other between the components normal to it. Then use the condition that the friction can not be greater than the normal force multiplied by the coefficient of static friction.

ehild
ehild
#9
Mar22-10, 02:13 PM
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Quote Quote by gaobo9109 View Post
From the answer, the range of value of a is from g(Ácosθ-sinθ)/(cosθ + Ásinθ) to gcotθ. So there must be two conditions for equilibrium. But from my two equations, i can only obtain one value of a. How do I find the other value of a for which the block is in equilibrium?
The other value of a (g cotθ) corresponds to that limit when the block falls vertically, just touching the ramp, but without interaction, so the normal force is zero. At even higher acceleration, the block would be above the ramp in the air during its free fall.

ehild


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