# Homework Help: Constant force problem

1. Sep 14, 2007

### BuBbLeS01

A block of mass 8.0 kg is pulled along a horizontal frictionless floor by a cord that exerts a force of 33.0 N at an angle 31.1° above the horizontal. What is the magnitude of the acceleration of the block?

Ok I am not sure how to go about solving this problem. I drew a diagram and I added weight and normal force. But now I don't know how to start the problem?

2. Sep 14, 2007

### PhanthomJay

What did you show for the applied force, try breaking it into x and y components.

3. Sep 14, 2007

### FedEx

I am here again. See as there is no friction on the surface the problem becomes much easier.Resolve the force in x and y components.The x comp will cause the motion in the block.

4. Sep 14, 2007

### BuBbLeS01

Thank you so much AGAIN!! You guys are awesome...I am understanding it more now!

5. Sep 14, 2007

### BuBbLeS01

Now I have to find the magnitude of the normal force. So I got the y component...so do I multiply that by mg?
y component= 33.0 sin 31.1

6. Sep 14, 2007

### FedEx

No there is no need for the y component as we donot need it. If there were friction then we would have to conisder it. Right now we have just one equation that is fcos31.1=ma

7. Sep 14, 2007

### PhanthomJay

The problem now asks to calculate the normal force. Why would you multiply? Consider all the forces acting in the y direction and use Newton 1 in the y direction to calculate the normal force. What forces act on the block in the y direction?

8. Sep 14, 2007

### BuBbLeS01

Weight acts in the y direction. I used that fcos31.1=ma equation but it says its wrong?

9. Sep 15, 2007

### PhanthomJay

Two parts to this problem if I understand correctly. The first asks you to calculate the acceleration of the block, which is in the x direction. Please show your work for that. You must brak up the applied force into its x componnet and then use F_net = ma.

The 2nd part acts you to calculate the normal force, which is the contact force that the floor exerts vertically up on the block, in the positive y direction. The weight of the block is another force that acts down on the block in the negative y direction, which you have identified. What is the other force that acts on the block in the y direction? Once you identify all three forces, then you apply Newton's law in the y direction to solve for the normal force. Hint: since the block is not moving in the y direction, what is its acceleraion in the y direction?

10. Sep 15, 2007

### Sabellic

My guess is as follows:

If the floor is frictionless, then neither Friction Force nor Normal Force need be considered. As well, the block moves only one direction horizontally, and there is no vertical motion. (The block doesn`t move up and down when it is pulled). Together, these two facts mean that the only force acting on the block is the tension (pulling) on the rope.

Therefore, to calculate the acceleration of the block, one need only to take the x component (in other words, the horizontal component) of the force into consideration.

11. Sep 15, 2007

### FedEx

Thats what i am trying to say.

12. Sep 15, 2007

### Sabellic

Oooops. LOL. Yep, I just looked at the above posts and you had already said the exact same thing before.