# Thread: Motion in an inclined plane

1. ## Motion in an inclined plane

A model car O is moving up an inclined road AB, at 30° to the horizontal,The weight W of the car is 1000N,the frictional force down the plane is 100N and the force on the car due to the engine is 800N up the plane AB.

So Calculate
(a)The component of the weight down the plane AB
(b)The resultant force on the car up the plane
(c)The reaction R on the car at right angles to the road AB if this is balanced by the component of the weight in the opposite direction OC.

2. ## Re: Motion in an inclined plane

Hi, srirahulan. My physics is a little rusty, but if I remember correctly this is more or less a trig problem. I have attached a picture that should get you started. A few other things that may be helpful:

1. You know what the green vector is from the given information.

2. The answer to part a) is the red part of the triangle (expressedin Newtons) in the picture.

3. To solve part b) take all the forces that are going "down the ramp" and subtract them from any forces that are known to be pushing "up the ramp"

4. The answer to part c) is the blue part of the triangle (expressed in Newtons) in the picture.

3. ## Re: Motion in an inclined plane

I can understand But I have some doubts in the (a) question i want to use sin 30 or cos 30.. Please Explain ...

4. ## Re: Motion in an inclined plane

For part a) you're trying to find the red part of the triangle, which is opposite to the 30 degree angle in the colored triangle. So we need to think of which trig function related the opposite side with the hypotenuse. Does this clear everything up for you?

5. ## Re: Motion in an inclined plane

I can't Understand please explain more.

6. ## Re: Motion in an inclined plane

Sine is the trig function that relates opposite with hypotenuse. So we want to use sine:

$\displaystyle \sin(30)=\frac{\text{red}}{1000}$.

Now solve for "red"

7. ## Re: Motion in an inclined plane

Okay Thanks.I can understand Now.