Problem on my Final. pulling box with rope problem

[J0hnnyD]

Homework Statement

A student pulled a box of books weighing 68kg by a rope at an angle of 58 degrees above the horizontal plane of the floor. The floor has a coefficient of kinetic friction of µk=.27. What is the acceleration of the box of books?

Homework Equations

∑F = ma
Force of friction = µk * Normal force

The Attempt at a Solution

I chose the direction the box of books was being pulled to be positive. Then I summed up all the forces in the x direction and applied Newton's Law F=ma and solved for a.

∑Fx = Force from rope in the x direction - Force of friction = ma
∑Fx = 68*cos(58) - .27*(68 * 9.81) = (68)a

solve for a.

This problem was on my final yesterday so the numbers arent correct but is my reasoning and approach to this problem right? I thought i knew how to do these types of problems but the answer i was getting wasnt a choice on the test.

Thanks in advance.

 

[tiny-tim]

Hi J0hnnyD!

The rope isn't horizontal, so the normal force will be less than mg.

 

[J0hnnyD]

Hi J0hnnyD!

The rope isn't horizontal, so the normal force will be less than mg.

I thought to get the force of friction its always µk*N. what is the normal force then? mg minus the y component of the force of the rope?

 

[novop]

Yea, exactly. The rope is pulling the box up a little, so the normal is actually less than mg.

 

[tiny-tim]

what is the normal force then? mg minus the y component of the force of the rope?

Yes …

the reason is that the acceleration in the normal direction is zero, so all the components of force in that direction must add to zero.

 

[jkerrigan]

I'm trying to understand this problem so if anyone could clarify.
But from what I am seeing here you really can't get a definite answer from
this problem because there is no force defined for the pull on the box,
so what you end up with is the Friction force depending on the pull in the
y direction which leads to unknown variables which depend on each other.
Somebody please explain to me if this is the case or I'm just not seeing something.

 

[tiny-tim]

Hi jkerrigan!

But from what I am seeing here you really can't get a definite answer from this problem because there is no force defined for the pull on the box,
so what you end up with is the Friction force depending on the pull in the
y direction which leads to unknown variables which depend on each other.

ah, but novop didn't have the exact question in front of him when he posted …

∑Fx = 68*cos(58) - .27*(68 * 9.81) = (68)a

solve for a.

This problem was on my final yesterday so the numbers arent correct …

… from the numbers in the posted (wrong) equation, we see that the pulling force was (by coincidence!) 68 N.