What are the forces acting on the crate on a frictionless ramp?

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Homework Help Overview

The discussion revolves around analyzing the forces acting on a crate being pushed up a frictionless ramp at a constant speed. The problem involves understanding the gravitational forces, applied forces, and the normal force in the context of a ramp inclined at 30 degrees.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants discuss the components of forces acting on the crate, including gravitational forces and applied forces. There is an exploration of the relationship between net force and acceleration, with some questioning the calculation of the normal force and its direction.

Discussion Status

The discussion is ongoing, with participants sharing their calculations and questioning the correctness of their approaches. Some have provided guidance on considering the net force and the implications of constant velocity, while others are verifying their calculations and seeking confirmation of their reasoning.

Contextual Notes

Participants are working under the constraints of a homework assignment, which may limit the information available and the methods they can use. There is a focus on ensuring accuracy in calculations, as some participants express concern about having limited attempts to submit their answers.

tjbateh
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Homework Statement


In the figure, a crate of mass m = 111 kg is pushed at a constant speed up a frictionless ramp (θ = 30°) by a horizontal force F . The positive direction of an x-axis is up the ramp, and the positive direction of a y-axis is perpendicular to the ramp. (a) What is the magnitude of F ? (b) What is the magnitude of the normal force on the crate?


Homework Equations





The Attempt at a Solution


For part A, I got 628.67 N which is correct.

I do not have part B, but I do have some other information that I found, which is correct.

Gravitational Force along x-axis is -544.455 N

Fx component of the applied force is 544.45 N

Fy component of the applied force is -314.341 N

Y component of the gravitational force is -943.023 N

Acceleration components are zero.

And the normal force of the crate is in the positive direction of the y-axis.
 
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Draw a free body diagram, put in all the forces and find what the normal force is. In this case it is not mg*cosθ, so don't use that. Remember that the acceleration in a direction perpendicular to the incline is zero.
 
would it just be 0? F=M*A...A=0, so F is equal to zero?
 
If by F you mean "the net force", in other words the sum of all the forces, then it is zero. That's because the problem is telling you that the velocity is constant. Constant velocity means zero acceleration which means zero net force according to Newton's Second Law.
 
ok so since the sum of all the forces is zero, could it be -628.67?
 
or should I find the magnitude of the gravitational force. And set F(g) + F + F(n) = 0?
 
In what direction is the normal force? The sum of all the components of forces in that direction must be zero. You know all the components of forces in that equation except for the normal force.
 
Does 1257.36 seem right?
I did (544.45i - 314.34j) + (-544.45i - 943.02j) + F(N) = 0

I get 0i + (-1257.36j) = 0

So F(N) is the Square Root of 1257.36^2, which is just 1257.36

I want to be completely sure of my answer, cause I only have one try left!
 
That looks about right if your components along the y-direction are correct - I didn't check them - but if you say they are, I believe you.
 

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