Newtons second law/friction problem

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The discussion revolves around calculating the force required to move a 68 kg crate, factoring in static friction and the angle of the pulling force. For part (a), the user initially calculated the static friction force incorrectly, leading to confusion about the applied force needed to overcome it. They realized that the pulling force has both vertical and horizontal components due to the 15-degree angle, which affects the normal force and, consequently, the friction. The correct approach involves breaking down the forces into components and applying Newton's second law accurately. Ultimately, the user acknowledges their misunderstanding and expresses gratitude for the clarification.
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Homework Statement


A 68 kg crate is dragged across a floor by pulling on a rope attached to the crate and inclined 15 degrees above the horizontal.

(a) If the coefficient of static friction is .50, what minimum force magnitude is required from the rope to start the crate moving?
(b) If \muk=0.35, what is the magnitude of the initial acceleration of the crate?

Homework Equations


F=ma

The Attempt at a Solution


part a:

I calculated the weight (9.8)(68 kg)=666N

then I got static friction: (666N)(.50)=333N

I substituted all the forces into Newtons second law and got:
ma=Normal+Gravity+Static+Applied force

after I plugged in the variables I got 0=333N + (force applied)(cosx)
(It's 0 because the box isn't moving so a=0)

when i solve for force i get 344.7N but the answer in the book is 300N
 
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The Force of the rope has 2 components.

The Y component is lightning the load of the 666 N by F*Sinθ . And it's F*Cosθ that is the force applied in the X direction to move it.

Try looking at what happens then if you account for that with θ = 15.
 
The force is unknown though. How can I find the components of the force if the magnitude is an unknown?
 
renob said:
The force is unknown though. How can I find the components of the force if the magnitude is an unknown?

But F is your only unknown.

Write out your equation and solve for F.

You know the components of F from θ = 15.
 
the original equation I had was 333N=Fcosx

would it be 333N=Fcosx + Fsinx?

sorry I'm very confused.
 
wait I see it now. thanks
 
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