# Homework Help: Horizontal Force Question

1. Nov 19, 2007

### jakecar

1. The problem statement, all variables and given/known data

A 30 kg chandelier hangs vertically from a 4 m long wire. What is the horizontal force necessary to displace the chandelier 0.10 m either way?

2. Relevant equations

F=ma

3. The attempt at a solution

I tried to figure out the angle made by the change in the wire's position if the chandelier moved .1 m. I eventually got that the angle should be about 1.43 degrees. From there, I have no idea what to do. I never learned how to figure how much force is necessary to move an object a certain distance. Thanks for any help!

2. Nov 20, 2007

### Bill Foster

Draw a free body diagram of the system when it is in the desired state. Label the forces and sum them up.

Since the object will not be moving in the final state, the net force must be zero. So you can calculate the the forces due to gravity and tension and they must be equal to the force you're looking for.

3. Nov 20, 2007

### jakecar

Ok so I set up a diagram. I have the weight of the chandelier going straight down (mg). I have the tension of the wire off at an angle in the first quadrant. I also have an unknown horizontal force that is perpendicular to the weight thats points to the right. Where does the length of the wire come in? I'm still kind of stuck at this point. Thanks for the help!

4. Nov 20, 2007

### Bill Foster

The forces in the y direction:

$$Tcos(\theta)-mg=0$$

The forces in the x direction:

$$Tsin(\theta)=F_x$$

You have to find $$\theta$$ using the length of the wire.

5. Nov 20, 2007

### jakecar

Ok I got the right answers using those equations for the forces in the x and y directions but how did you get those equations? I can't figure it out.... Thanks!

6. Nov 21, 2007

### Bill Foster

Sum up the forces.

When the chandelier is displaced 0.1m, the 4m rope is at some angle.

The forces acting on the chandelier can be broken into those along the x-axis and those along the y-axis.

Picture a right triangle. The hypotenuse is the rope and it's 4m long. The vertical side is where the rope would be if it were hanging straight down. And the base, or horizontal side, is 0.1m.

I'm assuming that the chandelier is not swinging or moving; just that there is a force acting on it to displace it 0.1m, and keep it there.

That force is denoted by $$F_x$$.

There has to be another force horizontally to counteract $$F_x$$, otherwise the mass would accelerate. The only other forces are the weight, which is $$mg$$, and the tension in the 4m rope, which is $$T$$.

$$mg$$ is perpendicular to $$F_x$$. But $$T$$ is at an angle $$\theta$$, which can be determined by the dimensions of the triangle.

Since $$T$$ is at an angle, it has both an x- and a y-component.

$$T$$'s x-component is $$Tsin(\theta)$$, and the x-component is $$Tcos(\theta)$$.

So now we have two forces along the y-axis, and two forces along the x-axis. Since the mass is stationary, the forces have to be equal.

y-axis: $$Tcos(\theta)=mg$$
x-axis: $$Tsin(\theta)=F_x$$

Solve for $$F_x$$

Dividing one equation by the other gives:

$$mgtan(\theta)=F_x$$

7. Nov 21, 2007

### jakecar

Thank you so so so much!