Force to push mass down inclined plane

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Homework Help Overview

The problem involves determining the force required to keep a 40 kg mass sliding down a frictionless inclined plane at a constant velocity, with the ramp set at an angle of 30 degrees.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the relationship between gravitational force and the motion of the mass on the incline, with one participant attempting to express the force in terms of the weight component along the ramp. Others suggest reviewing the forces acting on the mass and the importance of the normal force.

Discussion Status

The discussion includes various interpretations of the forces involved, with some participants providing insights into the calculations needed while others seek clarification on the underlying concepts. There is no explicit consensus reached, but guidance has been offered regarding the forces acting on the mass.

Contextual Notes

One participant notes a lack of understanding due to missed lessons, which may affect their ability to engage fully with the problem. The problem is framed within the constraints of a frictionless scenario.

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Homework Statement



The angle of the ramp is 30 degrees. The mass is 40 kg. What force must be applied to keep the mass sliding down the ramp at constant velocity. The ramp is frictionless*

Homework Equations


I have missed the past few lessons in class, so I'm not sure. I do have a bunch of equations, but I have little clue on how to use them for this problem

The Attempt at a Solution

 
Last edited:
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Assuming there is no friction, the force required to cause the mass to move at a constant rate is equal to the component of the mass's weight down the 30-degree incline. When this happens, the total net force on the mass is zero as the accelerative force (gravity) is canceled out, and it moves at a constant rate. The component of weight down the ramp is given by the expression mgcos(30). This, the force required is mgcos(30), or (40 kg)(9.8 m/s/s)(.866). It will be directed in the direction opposite the component of gravity down the ramp.
 
I got it, thanks guys
 
Last edited:

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