# Inclined plane

• February 19th 2008, 10:06 PM
DivideBy0
Inclined plane
Hi again, I've got a physics quesion here...

In the diagram, what is the force acting down the slope?

By trigonometry, I think it should be $\sin{\theta}=\frac{mg}{F}\Rightarrow F=\frac{mg}{\sin{\theta}}$, but according to the teacher it is $mg\sin{\theta}$. How?
• February 19th 2008, 10:51 PM
earboth
Quote:

Originally Posted by DivideBy0
Hi again, I've got a physics quesion here...

In the diagram, what is the force acting down the slope?

By trigonometry, I think it should be $\sin{\theta}=\frac{mg}{F}\Rightarrow F=\frac{mg}{\sin{\theta}}$, but according to the teacher it is $mg\sin{\theta}$. How?

You find 3 forces forming a parallelogram of forces:

1. the weight of the solid.
2. a force acting perpendicular on the surface of the inclined plane. This force is necessary to calculate the friction.
3. a force acting parallel to the surface of the inclined plane.

Since $|\theta| = |\theta_2|$ (you are dealing with similar right triangles) you'll see at once that your teacher is right.
• February 20th 2008, 12:37 AM
DivideBy0
Thanks earboth, but I guess what I really wanted to know is 'why doesn't it work?' It seems so intuitive to immediately perform trigonometry on $\theta$.
• February 20th 2008, 01:10 AM
earboth
Quote:

Originally Posted by DivideBy0
Thanks earboth, but I guess what I really wanted to know is 'why doesn't it work?' It seems so intuitive to immediately perform trigonometry on $\theta$.

The weight (on earth!) of the solid is calculated by:

$w = m \cdot g$

The weight is a force acting in the direction of the center of the earth that means perpendicular (or nearly perpendicular) to the surface of the earth.

This force is drawn in black in my sketch.

The force F (in red) (do you call it downhill force?) , the weight and the angle $\theta$ belong to a right triangle.

$\sin(\theta) = \frac{F}{w} = \frac F{m \cdot g}$