Mass on a friction less incline. What force is needed to stop the block from sliding.

[danpiz23]

Homework Statement

A block with mass 10kg rests on a smooth, friction less ramp that is inclined at an angle of 45 degrees with the ground. How much force must be applied in the direction parallel to the ground to prevent the block from sliding down the ramp?

IT TURNS OUT THAT THE MAGNITUDE OF THE FORCE THAT MUST BE APPLIED IN THE DIRECTION PARALLEL TO THE GROUND IN ORDER TO PREVENT THE BLOCK FROM SLIDING DOWN THE RAMP IS EQUAL TO THE WEIGHT OF THE BLOCK (IN NEWTONS). Explain why this is the care. Will this always be the case for a block on a friction less ramp?

Homework Equations

I found the answer (please check) as 69.3N, the teacher gave us a follow up page saying the force should be equal to the weight of the block in Newtons..Did I do this wrong?? The follow up page also asks to explain why the parallel force to the ground is equal to the weight of the block in Newtons. I don't get that.

The Attempt at a Solution

Weight = 10.9.8 = 98N
W vector <0,-98>
Vector u + Vector V = Vector W
F=-u
magnitude of F= magnitude of U
98 sin 45 = 69.3 N
A force of 69.3 N parallel to the plane will keep the weight from sliding.

The Attempt at a Solution

 

[PhanthomJay]

No, that is not correct. Draw a free body diagram and identify ALL the forces acting on the block. (There are 3). Break them into x and y components and use Newton 1 in both directions.

 

[danpiz23]

I know normal force and gravity are acting on the block, what else?? I drew the FBD, but I am confused on what else to do. I was following an example from our book done the same way.

 

[PhanthomJay]

I know normal force and gravity are acting on the block, what else?? I drew the FBD, but I am confused on what else to do. I was following an example from our book done the same way.

The gravity (weight) force acts straight down. The normal force acts perpendicular to the block pointing toward the block at the point of contact. Then there is the applied force acting parallel to the ground pushing toward the block. Find the x and y components of each force.

 

[rwisz]

Your calculation is correct. The magnitude of the force needed to prevent the block from sliding down the friction less incline is equal to the weight of the block. This is because in the absence of friction, the only force acting on the block is its weight, which acts in a vertical direction. The force needed to prevent the block from sliding down the incline must be equal in magnitude and opposite in direction to the component of the weight that is acting parallel to the incline. This component is equal to the weight of the block multiplied by the sine of the incline angle, which is 45 degrees in this case. Therefore, the magnitude of the parallel force must be equal to the weight of the block, which is 69.3 N in this case.

This will always be the case for a block on a friction less incline, as long as the incline is not changing and there are no external forces acting on the block. In this scenario, the weight of the block is the only force acting on it, and the magnitude of the parallel force needed to prevent sliding will always be equal to the weight of the block. This is because, in the absence of friction, the weight of the block is the only force that can cause the block to accelerate down the incline.