What is the minimum force at this angle that will move the box?

[zehkari]

Homework Statement

[/B]

2. Homework Equations

Fs = Nμ
W = mg
Fy = Fsin(θ)
Fx = Fcos(θ)

The Attempt at a Solution

[/B]
Hello all at PhysicsForums,

I have a question on angled force and static friction. From my attempt below, I think I have found the maximum force required and would need to get the minimum. Then using trigonometry I could derive sec(θ) and tan(θ).

Have tried some algebra to get to the quation in the question, but I have been unsuccessful.

If anyone could help me that would be great,
Thanks.
Zehkari.

 

[gneill]

Hint: How might that cos(θ) in the denominator become a 1?

 

[zehkari]

Something with using the identity
cos²(θ)+sin²(θ) = 1
so
1 - sin²(θ) = cos²(θ)

 

[gneill]

Something with using the identity
cos²(θ)+sin²(θ) = 1
so
1 - sin²(θ) = cos²(θ)

Nope. even simpler.

 

[zehkari]

By dividing by itself? -_-

 

[Chandra Prayaga]

By dividing by itself? -_-

Try that!

 

[zehkari]

Try that!

Dividing the euqation by cos(θ), I get:

F/cos(θ) = mg.μ/1+μ.tan(θ)

so,

F = mg.μ.cos(θ)/1+μ.tan(θ)

 

[gneill]

I just noticed that you have a sign error in the denominator of your solution attempt. The μ_ssin(θ) term should have ended up negative. You'll have to go through your work again to find out where the sign error occurred.

Dividing the euqation by cos(θ), I get:

F/cos(θ) = mg.μ/1+μ.tan(θ)

No, check your algebra. I'd suggest manipulating only the right-hand side, leaving the left alone. Divide the numerator and denominator by cos(θ) (hence effectively multiplying the right-hand side by 1).

 

[Chandra Prayaga]

Dividing the euqation by cos(θ), I get:

F/cos(θ) = mg.μ/1+μ.tan(θ)

so,

F = mg.μ.cos(θ)/1+μ.tan(θ)

Please check your work, before this last step. For example, check the line you wrote for F cosθ = (mg - F sinθ) μ_s

 

[zehkari]

Thank you gneill and Chandra Prayaga, makes sense now. Apologies for the silly error on the sign.

 

[gneill]

Thank you gneill and Chandra Prayaga, makes sense now. Apologies for the silly error on the sign.

I hope that you've reached the desired result? Otherwise, please post your work and we'll be happy to offer more help.

 

[kuruman]

You'll have to go through your work again to find out where the sign error occurred.

The error is in the diagram that shows N=mg - μ_kF sinθ. Because the external force is pushing down and across, the reaction force should be greater than mg, not less.

 

[zehkari]

I hope that you've reached the desired result? Otherwise, please post your work and we'll be happy to offer more help.

I think its good now. As the numerator is divided by cos(θ), you get sec(θ). Then as the denominator is divided by cos(θ), you get 1-μ.tan(θ). Thank you for your help.

The error is in the diagram that shows N=mg - μ_kF sinθ. Because the external force is pushing down and across, the reaction force should be greater than mg, not less.

Thank you, yes I have it as greater than now.