What Force Stops a 10kg Block from Sliding on a 45-Degree Frictionless Ramp?

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Homework Help Overview

The problem involves a 10kg block resting on a frictionless ramp inclined at 45 degrees. The original poster seeks to determine the force required to prevent the block from sliding down the ramp and questions whether this force is equal to the weight of the block.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants discuss the forces acting on the block, including weight and normal force, and the need to analyze these forces using a free body diagram. There is uncertainty about the relationship between the applied force and the weight of the block.

Discussion Status

Some participants have provided guidance on drawing a free body diagram and breaking forces into components. There is ongoing exploration of the forces involved, but no consensus has been reached regarding the correct approach or solution.

Contextual Notes

The original poster references a follow-up page from the teacher that suggests the applied force should equal the weight of the block, raising questions about the generality of this statement for blocks on frictionless ramps.

danpiz23
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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

 
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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.
 


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.
 


danpiz23 said:
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.
 

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