Accel of Crate Pushed at 30° Below Horizon

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Homework Help Overview

The discussion revolves around determining the acceleration of a crate being pushed at an angle of 30 degrees below the horizontal. The problem involves a mass of 2.0 kg, an applied force of 10 N, and a coefficient of kinetic friction of 0.047.

Discussion Character

  • Mixed

Approaches and Questions Raised

  • Participants discuss the setup of free body diagrams and the application of force components in the x and y directions. There are questions regarding the correct expression of forces and the role of friction in the calculations. Some participants suggest expressing forces in terms of magnitudes for clarity.

Discussion Status

The discussion is active, with participants providing feedback on each other's setups and clarifying points of confusion. There is a focus on ensuring that the contributions of gravity and friction are accurately represented in the equations. While some participants express confidence in the setups, others raise questions about the signs and components of the forces involved.

Contextual Notes

Participants are navigating through potential typos and misunderstandings regarding the directionality of forces, particularly the frictional force in relation to the applied force. There is an emphasis on maintaining clarity in vector notation and the implications of the angle of application.

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Homework Statement



Find the acceleration of the crate when it is pushed by the same force at an angle of 30 degree below the horizon.

Given mass = 2.0kg Applied Force = 10 N, and uk = 0.047

Homework Equations



Fx/y = ma = sum of all force vectors
Fk = ukN

The Attempt at a Solution



So I did a free body diagram. Here is my attempt

<Fx = ma
<Fy = 0

Fx = ma = Fcos(-30) * Fk
Fy = 0 = Fsin(-30)+N-mg

Is that it? Anything tricky?

I just want to make sure all my setups are correct.
 
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That's right, but to prevent confusing yourself, you might want to express everything in terms of magnitudes. So ma=Fcos(30)*Fk: the magnitude of ma is equal to the magnitude of Fcos(-30)*Fk. N=mg+Fsin(30), because the ground pushes up against both gravity and the y component of the force F. I find that this approach is much more intuitive than using Fcos(-30) and Fsin(-30).

EDIT: Oops, that's not right. You forgot to account for the contribution of gravity to the normal force.
 
jwxie said:

Homework Statement



Find the acceleration of the crate when it is pushed by the same force at an angle of 30 degree below the horizon.

Given mass = 2.0kg Applied Force = 10 N, and uk = 0.047

Homework Equations



Fx/y = ma = sum of all force vectors
Fk = ukN


The Attempt at a Solution



So I did a free body diagram. Here is my attempt

<Fx = ma
<Fy = 0

Fx = ma = Fcos(-30) * Fk
Fy = 0 = Fsin(-30)+N-mg

Is that it? Anything tricky?

I just want to make sure all my setups are correct.
Why are you multiplying the x-component of the applied force by the force of friction?

ideasrule said:
EDIT: Oops, that's not right. You forgot to account for the contribution of gravity to the normal force.
jwxie said:
Fy = 0 = Fsin(-30)+N-mg
didn't he do that here?
 
yeah that was a typo. it was suppose to be minus
Fx = F - Fk
and Fk also has x, y components, which is along the x axis
 
Last edited:
jwxie said:
yeah that was a typo. it was suppose to be minus
Fx = F - Fk
and Fk also has x, y components, which is along the x axis

Looks good to me. Just make sure you keep your signs straight -- I'd write F + Fk with Fk being a negative vector to denote direction. I do, however, believe you meant the right thing.
 
xcvxcvvc said:
Looks good to me. Just make sure you keep your signs straight -- I'd write F + Fk with Fk being a negative vector to denote direction. I do, however, believe you meant the right thing.

Hi, thanks for your reply.

I am sorry, but why is it F + Fk?

My thought would be that Fk is moving in the opposite direction of Fx.

Or you meant let Fk = -(uk*N)
So in general, the vector sum F + Fk
 

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