How Do You Calculate Force on an Inclined Plane with and without Friction?

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Homework Help Overview

The discussion revolves around calculating the force required to push a 90 kg box up an inclined plane at a 28° angle, both with and without friction. The original poster expresses confusion regarding the application of trigonometric functions in the context of force decomposition and the role of friction in the calculations.

Discussion Character

  • Exploratory, Conceptual clarification, Problem interpretation, Assumption checking

Approaches and Questions Raised

  • Participants discuss the need for vector decomposition of forces acting on the box, particularly the gravitational force. Questions arise about the reasoning behind using sine functions and the interpretation of the normal force and friction in the context of the problem.

Discussion Status

Participants are actively engaging with the problem, offering suggestions to draw diagrams and focus on the components of forces. There is an emphasis on understanding the underlying concepts rather than simply applying formulas. No consensus has been reached, but there is a productive exploration of ideas and clarifications being sought.

Contextual Notes

The original poster notes a lack of instruction on kinetic friction in their course, which adds to their uncertainty regarding part (b) of the problem. There is also a mention of needing to visualize the forces involved, indicating a potential gap in understanding the physical setup.

questionable
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Homework Statement



A 90 kg box is pushed by a horizontal force F at constant speed up a ramp inclined at 28°, as shown. Determine the magnitude of the applied force:
a. when the ramp is frictionless.
b. when the coefficient of kinetic friction is 0.18.

Homework Equations



F= ma, Ffr = μk * Fn

The Attempt at a Solution



Everywhere I look online I keep getting told that for part (a) I need to do sin28 * 882N (i.e. sin28 * mg) but I have NO idea why. Could somebody draw me a picture of the vector decomposition because I really really do not understand why you'd multiply.

For part (b) my course hasn't even taught us about kinetic friction yet, but through looking around I've found out that Ffr = μk * Fn. So once I've found the force of friction, what do I do with it?

Thanks for any help...
 
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questionable said:

Homework Statement



A 90 kg box is pushed by a horizontal force F at constant speed up a ramp inclined at 28°, as shown. Determine the magnitude of the applied force:
a. when the ramp is frictionless.
b. when the coefficient of kinetic friction is 0.18.

Homework Equations



F= ma, Ffr = μk * Fn

The Attempt at a Solution



Everywhere I look online I keep getting told that for part (a) I need to do sin28 * 882N (i.e. sin28 * mg) but I have NO idea why. Could somebody draw me a picture of the vector decomposition because I really really do not understand why you'd multiply.

For part (b) my course hasn't even taught us about kinetic friction yet, but through looking around I've found out that Ffr = μk * Fn. So once I've found the force of friction, what do I do with it?

Thanks for any help...

Have you tried drawing a diagram?

If you upload a picture of your attempt I can see where you're having the issue.

The key to these questions are good force diagrams, so having one drawn for you won't solve anything in the long run.
 
BOAS said:
Have you tried drawing a diagram?

If you upload a picture of your attempt I can see where you're having the issue.

The key to these questions are good force diagrams, so having one drawn for you won't solve anything in the long run.

This is the closest guess that I have. I don't have a clue really. Sorry for the bad drawing.
 

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questionable said:
This is the closest guess that I have. I don't have a clue really. Sorry for the bad drawing.

Just focus on the force of gravity for a moment. Draw it so that the force of gravity is the hypotenuse and the two legs are parallel and perpendicular to the ramp. (This way you've broken up the force of gravity into the components perpendicular to the ramp and parallel to the ramp.)

Then just think about the frictionless case. If the box is kept at a constant velocity, then what must be true? How can you make this true?
 
Okay, I've done that. What are the components of the gravity vector though? Isn't it just pointing down?
 
questionable said:
Okay, I've done that. What are the components of the gravity vector though? Isn't it just pointing down?

The weight of the box does point straight down, but think about the direction of the normal force, and you should see that you need to decompose mg into two components.
 
questionable said:
Okay, I've done that. What are the components of the gravity vector though? Isn't it just pointing down?
Yes, but the coordinate system imposed in the solution is one in which the axes are parallel and perpendicular to the plane of the incline. In this case, you can decompose the gravity vector along these mutually orthogonal directions.
 

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