Webpage title: Optimal Angle for Pulling a Crate with a Friction Rope

[yolo123]

A rope is tied to a large crate, which is sitting on a flat surface. The coefficient of static friction between the crate and the ground is 0.9. If a person is to pull on the rope with the minimum force needed such that the crate begins to slide, the angle between the rope and the ground should be
A)

greater than 0 degrees but less than 90 degrees
B)

0 degrees (rope is horizontal)
C)

90 degrees I know the answer is A. I found this question in practice problems for midterms.

Now, I was trying to prove it:
horizontal pulling: F=ukmg.

greater than 0 degrees but less than 90 degrees:
Horizontal component: uk(mg-sinthetaF) (Let F be magnitude of force.)
Vertical component: costhetaF
Total magnitude:
(Pythagoras) and (sin^2theta+cos^2theta=1)
0.81mg-1.8sinthetaF+F^2=Magnitude.

How do I prove 0.81mg-1.8sinthetaF+F^2 > 0.9mg?

 

[haruspex]

Horizontal component: uk(mg-sinthetaF) (Let F be magnitude of force.)
Vertical component: costhetaF
Total magnitude:
(Pythagoras) and (sin^2theta+cos^2theta=1)
0.81mg-1.8sinthetaF+F^2=Magnitude.

That last expression is incorrect. You seem to have dropped some squarings.
If you had the right expression you could equate that to F and simplify.
A more direct method is just to write out the statics equations in the vertical and horizontal directions.

 

[paisiello2]

What is "Magnitude"?

Try summing the forces in each direction.

 

[yolo123]

Hi, I don't really understand what you are saying. Could you give more hints?

 

[FermiAged]

The net normal force is mg-Fsin(theta).

So Horizontal force balance is: F cos(theta)-u(mg - Fsin(theta)) = 0

Collecting terms, F(theta) = umg/(cos(theta)-usin(theta)

dF/d(theta) = (umgsin(theta)+u^2mgcos(theta))/(cos(theta)-usin(theta))^2

Setting the derivative to zero gives me an angle of 42 deg. [arctan (0.9)]

Do check my math!

 

[haruspex]

The net normal force is mg-Fsin(theta).

So Horizontal force balance is: F cos(theta)-u(mg - Fsin(theta)) = 0

Collecting terms, F(theta) = umg/(cos(theta)-usin(theta)

dF/d(theta) = (umgsin(theta)+u^2mgcos(theta))/(cos(theta)-usin(theta))^2

Setting the derivative to zero gives me an angle of 42 deg. [arctan (0.9)]

Do check my math!

Hi FermiAged,

This is a Homework Forum. The idea is to provide hints and point out errors, not provide solutions. Please see the guidelines.

 

[FermiAged]

Sorry. Iam new at this and got carried away.

 

[haruspex]

Hi, I don't really understand what you are saying. Could you give more hints?

Do you understand how to draw a free body diagram? How to find the vertical and horizontal components of forces? How to write out the statics equations ∑F=ma=0?

 

[yolo123]

YES! Do not worry. I solved this problem a few hours ago. Sorry I forgot to give you feedback! I'm stuck on another problem for which I posted another thread. That one is much more conceptual than this one :S