Finding tention of rope, accelerating box.

In summary, the problem involves a 500N box being pulled by a rope at a 30 degree angle, with a kinetic frictional coefficient of 0.4. The box accelerates at 1.2 m/s^2 horizontally and the task is to find the tension of the rope. Using Newton's second law, the sum of forces on the y axis is found to be 0, and the sum of forces on the x axis is found to be Tcos(30°) - 0.4(500N) = (51kg)(1.2m/s^2). This results in a tension of 302N. However, the normal force should not be assumed to be simply the weight of the
  • #1
guipenguin
7
0
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
 
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  • #2
Because the rope is at an angle, the normal force is not simply the weight of the box!
 
  • #3
Nabeshin said:
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
 
  • #4
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!
 

Related to Finding tention of rope, accelerating box.

1. What is tension of a rope?

Tension is a force that is transmitted through a rope or string when it is pulled tight. It is a pulling force that is directed along the length of the rope or string.

2. How do you calculate tension in a rope?

Tension can be calculated using the formula T = m x a, where T is the tension, m is the mass of the object being pulled, and a is the acceleration of the object.

3. How do you find the acceleration of a box?

The acceleration of a box can be found by dividing the net force acting on the box by its mass, or by using the formula a = F/m, where a is the acceleration, F is the net force, and m is the mass of the box.

4. What factors affect the tension of a rope?

The tension of a rope can be affected by various factors, such as the weight of the object being pulled, the angle at which the rope is pulled, and the type and thickness of the rope.

5. How can I increase the tension in a rope?

The tension in a rope can be increased by increasing the force applied to the rope, decreasing the angle at which the rope is pulled, or using a stronger and thicker rope.

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