Understanding Static Friction on a Ramp: Why is the Object Stationary?

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The discussion clarifies that an object on an inclined plane remains stationary when the static friction force equals the component of its weight acting down the ramp. The maximum static friction force, calculated using the normal force and the coefficient of static friction, represents the upper limit but does not dictate the actual friction force. The actual static friction force adjusts to match the weight component down the ramp, ensuring equilibrium. This means that while the static friction can be greater, it only needs to be equal to the weight component to keep the object at rest. Thus, the object does not move because the forces are balanced.
EF17xx
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Hi ,

The Static Friction force of an object on a ramp(inclined plane) making an angle with the horizontal is greater than the weight component down the ramp. My textbook says that this means that the object on the ramp is stationary as the frictional force is greater than the weight component force down the ramp. However for an object to be still or in equilibrium shouldn't the forces add up to 0. In this case if the frictional force is greater, the object should technically be moving up the ramp and intuitively this is obviously wrong so could someone please clarify this for me ? as I am a bit confused..
 
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The confusion here is due to an unclear problem statement. You calculate the maximum possible friction force from the normal force and friction coefficient. You use the FBD to find the actual friction force.
 
EF17xx said:
The Static Friction force of an object on a ramp(inclined plane) making an angle with the horizontal is greater than the weight component down the ramp.
How do you know this? I suspect you think that the force of static friction is calculated from fs = μsN (N = normal force). This is not the correct way to find the force of static friction. As @jrmichler suggests, you should draw the FBD and figure out what it is.

The right side of the equation fs = μsN is the largest value that the force of static friction can have. This means that static friction can be no larger than that value. If the component of the weight down the ramp is less than that maximum value, then the object will be at rest on the incline.
 
Ok so the force of static friction will not actually be the largest value and will only be big enough to cancel out the component down the ramp at the points on the ramp that it can but as we know that the force of static friction can be larger than the component down the ramp at that specific point as I calculated both the max force of static friction and the size of the component down the ramp they will be cancelling each other out. But the magnitude of the force of static friction will simply be equal to the magnitude of the force component down the ramp and therefore the object is at rest. Is this correct?
 
EF17xx said:
But the magnitude of the force of static friction will simply be equal to the magnitude of the force component down the ramp and therefore the object is at rest. Is this correct?
That is absolutely correct.
 
For simple comparison, I think the same thought process can be followed as a block slides down a hill, - for block down hill, simple starting PE of mgh to final max KE 0.5mv^2 - comparing PE1 to max KE2 would result in finding the work friction did through the process. efficiency is just 100*KE2/PE1. If a mousetrap car travels along a flat surface, a starting PE of 0.5 k th^2 can be measured and maximum velocity of the car can also be measured. If energy efficiency is defined by...

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