Horizontal force with a crate at a 27 degree angle

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Homework Help Overview

The discussion revolves around a physics problem involving a 100 kg crate being pushed up a frictionless ramp inclined at 27 degrees. Participants are tasked with determining the horizontal force required to maintain constant speed and the normal force exerted by the ramp on the crate.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants discuss the relationship between forces acting on the crate, including gravitational force and the components of the applied force. There are attempts to use trigonometric relationships to resolve forces into components. Questions arise regarding the lack of information and the correct interpretation of the applied force's direction.

Discussion Status

The discussion is ongoing, with participants exploring different interpretations of the problem. Some guidance has been offered regarding the components of forces, but there is no consensus on the correct approach or sufficient information to proceed. Confusion remains about the application of forces and the setup of equations.

Contextual Notes

Participants note the absence of a figure that is crucial for visualizing the problem. There is also a recognition of the challenge posed by the need to solve for multiple unknowns without complete information.

bearhug
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A 100 kg crate is pushed at constant speed up the frictionless 27° ramp shown in the figure. What horizontal force F is required?
What force is exerted by the ramp on the crate?

I can't paste the figure up so I hope this question describes the scenario well enough.

What's giving me problems is how am I suppose to find F when I don't even know n? I figured I use the angle to help solve for others. I know Fg=mg which is (100)(9.8) right? Would that give me one side of the triangle?
 
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The problem says at a constant speed, therefore a=0. Also, there is no friction slowing you down. F=ma, a=0, so the net force is zero. You need to find the vertical and horizontal components of the applied force. So F(applied)*cos(27)=F(applied,horizontal), and F(applied)*sin(27)=F(applied, Vertical).

That should get you off to a good start, remember you don't know F(applied), so you'll have to solve for two variables (substitution in otherwords).
 

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I'm still confused. First of all did you post a figure cause it's not showing up. I understand that I need to use substitution to solve for one and then solve for the other but I still feel like I don't have enough information. This is what I have down:

Xcomponent
∑Fx= max
mgsin27 = ma
a=gsin27

Ycomponent
∑Fy= may
N-mgcos27= 0
N= mgcos27

Is this at all right?
 
In the y-sum you forgot to add F, so you have mgcos27 + Fsin27 = N.
 
bearhug said:
A 100 kg crate is pushed at constant speed up the frictionless 27° ramp shown in the figure. What horizontal force F is required?
Is the applied force horizontal? or parallel to the ramp?
 

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