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

In summary, the author attempted to solve for the maximum force that can be exerted using static friction, and found that it is mg.μ.cos(θ)/1+μ.tan(θ).
  • #1
zehkari
22
3

Homework Statement


[/B]
Question.JPG


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.

attempt.JPG
 

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  • #2
Hint: How might that cos(θ) in the denominator become a 1?
 
  • #3
Something with using the identity
cos2(θ)+sin2(θ) = 1
so
1 - sin2(θ) = cos2(θ)
 
  • #4
zehkari said:
Something with using the identity
cos2(θ)+sin2(θ) = 1
so
1 - sin2(θ) = cos2(θ)
Nope. even simpler.
 
  • #5
By dividing by itself? -_-
 
  • #6
zehkari said:
By dividing by itself? -_-
Try that!
 
  • #7
Chandra Prayaga said:
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)
 
  • #8
I just noticed that you have a sign error in the denominator of your solution attempt. The μssin(θ) term should have ended up negative. You'll have to go through your work again to find out where the sign error occurred.
zehkari said:
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).
 
  • #9
zehkari said:
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) μs
 
  • #10
Thank you gneill and Chandra Prayaga, makes sense now. Apologies for the silly error on the sign.
 
  • #11
zehkari said:
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.
 
  • #12
gneill said:
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 - μkF sinθ. Because the external force is pushing down and across, the reaction force should be greater than mg, not less.
 
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  • #13
gneill said:
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.

kuruman said:
The error is in the diagram that shows N=mg - μkF 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.
 
  • Like
Likes gneill

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

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

The minimum force required to move a box at a specific angle depends on various factors such as the weight of the box, the coefficient of friction between the box and the surface, and the angle at which the force is applied. Therefore, it is not possible to determine the exact minimum force without knowing these factors.

What is the coefficient of friction and how does it affect the minimum force?

The coefficient of friction is a measure of the resistance between two surfaces when they are in contact with each other. It is a dimensionless value and is affected by factors such as the type of surface, the roughness of the surface, and the force applied. A higher coefficient of friction means that more force is required to move the box.

Can the minimum force at an angle be calculated using a formula?

Yes, there are various formulas that can be used to calculate the minimum force required to move a box at a specific angle. These formulas take into account the weight of the box, the coefficient of friction, and the angle at which the force is applied. However, it is important to note that these formulas provide an estimate and the actual force required may vary due to other factors.

How can the minimum force at an angle be measured experimentally?

The minimum force required to move a box at a specific angle can be measured experimentally by setting up a controlled experiment. This may involve using a force meter to measure the force applied at different angles, varying the weight of the box, and changing the coefficient of friction by using different surfaces. The results of these experiments can help determine the minimum force at a particular angle.

What are some practical applications of knowing the minimum force at an angle?

Knowing the minimum force required to move a box at a specific angle can be useful in various practical applications such as designing machines and equipment, determining the capabilities of different surfaces, and understanding the mechanics of how objects move. This knowledge can also help in optimizing the use of force and reducing the risk of injury when handling heavy objects at different angles.

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