Question

A block of mass 9 kg rest on a plane msking an angle of 16 degree with horizontal. determine component of weight normal to the plane.​

Answer 1

Explanation:

A block of mass 9 kg rest on a plane msking an angle of 16 degree with horizontal. determine component of weight normal to the plane.

answer is 24.33

Answer 2

Answer: As the determining component of weight normal to the plane 24.33 N

Explanation: In a free body diagram, the net forces along X-axis and Y-axis balance each other.

The weight of the body acts along the vertical downwards direction.

The formula for the weight of the body is:

$W=M*g$

Where M is the mass of the body.

            g is the gravity.

As the mass of the block,  M=9 kg

          gravity,  g=10 $m/s^{2}$

          angle with horizontal, θ =160°

As a Component of weight normal to plane = Mgcosθ

So, the component of weight normal to the plane = (9 kg)*(10 m/s²) cos160 = 84.86.

As the determining component of weight normal to the plane 84.86.

=Mgsin16°

=88.29×0.27

=24.33

It is a component of weight normal to the plane 24.33

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