Finding tention of rope, accelerating box.

[guipenguin]

A 500N box is pulled is pulled by a rope at a 30 degree angle. Box accelerates at 1.2 m/s^2, horizontally, with kinetic frictional coefficient = 0.4 Find tension of rope.

Now, I'd have to be using Newton's second law, here.

Since there is no acceleration on the y axis: sum of forces_y = ma = 0

Now, as for the forces on the x axis, I thought I would need to do something like this, to find the tension of the rope. Let T = tension of rope.

As the box weights 500N, and although g was not defined, I'd have to guess that this was on Earth and that mass = 500N / (9.8m/s^2) = 51kg.

sum of forces_x = Tcos(30°) - 0.4(500N) = (51kg)*(1.2m/s^2)

T = 302N Would this be correct?

Thanks,
John

 

[Nabeshin]

Because the rope is at an angle, the normal force is not simply the weight of the box!

 

[guipenguin]

Because the rope is at an angle, the normal force is not simply the weight of the box!

What does the normal force have to do with this problem? Although the rope is at an angle, the box remains parallel with the x-axis, so f_gravity = 500N would be equivalent, but opposite direction, to the normal, but again I don't see how this applies in the context of my problem.

If you could please elaborate, that would be great.
John

 

[Nabeshin]

I mention the normal force because the frictional force is equal to the product of the normal force and coefficient of friction. You used the normal force to be merely the weight of the box, which is not the case. You seem comfortable saying the rope exerts a force of Tcos(30) in the x, but it also exerts a force in the y!