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# What is normal rection if a weight of 600kg is kept on tha inclined plane at 30sec.

Piyush Kumar Behera
436 Points
22 days ago
Assuming the question is about an incline plane at 30 degrees.
We know that the weight is directed downwards of magnitude mg
We can split this weight into two components, one towards the normal of the inclined plane and other parallel to the inclined plane.
The component of weight along the normal to the inclined plane would be mg cos$\Theta$.
As there is no net force along the normal. Hence the normal is mg cos$\Theta$ = 600 x 9.8 x ($\sqrt{3}$/2)