Accel of Crate Pushed at 30° Below Horizon

[jwxie]

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.

 

[ideasrule]

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.

 

[xcvxcvvc]

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?

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

Fy = 0 = Fsin(-30)+N-mg

didn't he do that here?

 

[jwxie]

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

 

[xcvxcvvc]

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.

 

[jwxie]

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