next up previous
Next: Limits Up: Tension on an inclined Previous: Draw free body diagrams

F=ma in component form

Now we need to figure out a good coordinate system to use and apply tex2html_wrap_inline669 for both the x and y components.

Funnily enough, the best coordinate system to use is the one tilted along the inclined plane as shown in the figure below.

  figure116

This will take some practice to figure out. You can use any coordinate system you want, but its good to use a simple choice as the answers tend to come out a lot more easily that way. With this choice you can see that the only vector that we have any trouble decomposing is the weight. The normal force and the tension lie in the direction of the y and x axes respectively. This way we have less trigonometry to do.

So decomposing the weight we have

  equation120

Because the mass is not moving, the acceleration of it is zero, so from the second law, that means that the net force acting on the block is zero.

equation123

This is a vector equation. We want to write it now at component form. (Remember my advice!).

In the y direction, the tension is zero, but the normal force is entirely in this direction. so

equation125

This says that

  equation127

In the x direction, the normal force is zero, but the tension entirely lies along this direction. Therefore

equation130

Therefore

  equation132

Now as we said above, from the third law, this means that the scale will read the same thing.

Remembering tex2html_wrap_inline681 , we have deduced that the scale will read tex2html_wrap_inline683 .



Joshua Deutsch
Wed Jan 7 17:12:17 PST 1998