You need to move a 500 N box across the floor. You pull on the box

[Jkblackbear08]

You need to move a 500 N box across the floor. You pull on the box with a rope which is also horizontal. The floor is not frictionless. You move the box so that it moves with a constant v of 3m/s by applying a 200 N force.

Coefficient of Friction?

Also,

same question except pull the box at an angle of 30 degrees from the horizontal, use 0.4 coefficient.

what is the tension in the rope?

I just really can't even get started on this one, i have no idea, my professor really didnt cover to much of this but gave us 4 similar questions on it, any help would help me out...

 

[cepheid]

This should have been posted in the homework help sub-forum using the associated template for homework threads (which would have provided a systematic framework for you to articulate and approach the problem).

The basic principle you need to understand in order to solve this problem is Newton's Second Law of Motion. Since the box's velocity is constant (in the horizontal direction) and zero (in the vertical direction), it is not accelerating, which implies that the forces on it are balanced (in both the horizontal and vertical directions). Therefore the NET horizontal force on the box must be zero. Since there are only two horizontal forces, the applied force due to the tension in the rope (forwards), and friction (backwards), these two must be equal to each other (in magnitude)

Therefore, you KNOW what the magnitude of the frictional force must be in this situation. You also know what the normal force is (because you are given the weight and you know that the vertical forces are balanced). So, you have enough information to determine the coefficient of friction for sliding motion between the box material and floor material.

 

[jack action]

Friction force = Coeff. friction X Normal force

The normal force is perpendicular to the friction force (the one pressing on the ground)

When there is a slope, the weight of the box is still pointing down and can be divided in two components: A horizontal one and a vertical one (with respect to the slope). Determine the friction force and carefully add the force horizontally (wrt the slope) to find the tension.

 

[cepheid]

For the second part of the problem, the applied force (or tension) is now NOT entirely horizontal. It has both horizontal and vertical components. You can figure out what these components are in terms of the unknown quantity (the tension), because you know the angle at which this force is applied.

Then it's just the same thing again. Newton's second law requires that if the acceleration is zero (horizontally and vertically) then:

- the sum of the forces in the horizontal direction is zero

- the sum of the forces in the vertical direction is zero

The balance of the forces in the vertical direction is a little different from in the first part of the problem, because instead of just two vertical forces (normal force and weight), there are now three (the third being the upward component of the tension). This means that the normal force will change -- it will not be the same as it was in the first part of the problem.

 

[pmacias]

I would approach this problem by first understanding the concept of friction and how it affects the movement of objects. Friction is a force that opposes motion and is caused by the interaction between two surfaces. In this case, the floor and the box are the two surfaces.

To calculate the coefficient of friction, we need to know the force of friction and the normal force (the force exerted by the floor on the box). In this scenario, the normal force would be equal to the weight of the box, which is 500 N. Therefore, the coefficient of friction can be calculated as:

Coefficient of friction = Force of friction / Normal force

Since the box is moving at a constant velocity of 3m/s, we can assume that the force of friction is equal to the applied force of 200 N. Therefore, the coefficient of friction would be:

Coefficient of friction = 200 N / 500 N = 0.4

This means that the coefficient of friction for this scenario is 0.4.

In the second scenario, where the box is pulled at an angle of 30 degrees from the horizontal, the tension in the rope would be different. This is because the force of tension is now acting at an angle and not in the same direction as the motion of the box. To calculate the tension in the rope, we can use the following formula:

Tension = Applied force / cosθ

Where θ is the angle between the applied force and the direction of motion. In this case, θ is 30 degrees. Therefore, the tension in the rope would be:

Tension = 200 N / cos30° = 200 N / 0.866 = 230.9 N

I hope this helps you understand the concept better. It is important to understand the fundamentals of friction and how it affects the movement of objects in order to solve problems like these.