a 50 kg block is placed on an inclined plane with an angle of 30. then find the components of the weight (1)perpendicular (2)parallel to the inclined plane.

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To find the components of the weight of a 50 kg block placed on an inclined plane with an angle of 30 degrees, we can use trigonometry.

1) The perpendicular component of the weight is the force acting perpendicular to the inclined plane. This is equal to the weight of the object multiplied by the cosine of the angle of inclination.

Given:
Weight (W) = 50 kg

Perpendicular component (P) = W * cos(angle)

P = 50 kg * cos(30 degrees)
P = 50 kg * √3/2
P ≈ 43.3 kg (rounded to one decimal place)

So, the perpendicular component of the weight is approximately 43.3 kg.

2) The parallel component of the weight is the force acting parallel to the inclined plane. This is equal to the weight of the object multiplied by the sine of the angle of inclination.

Parallel component (Q) = W * sin(angle)

Q = 50 kg * sin(30 degrees)
Q = 50 kg * 1/2
Q = 25 kg

So, the parallel component of the weight is 25 kg.

Therefore, the components of the weight are:
1) Perpendicular component ≈ 43.3 kg
2) Parallel component = 25 kg

think of a weight w on a horizontal plane (θ=0). All of the force is downward. That means that

perpendicular is w*cosθ
parallel is w*sinθ

As the board is tilted, less and less is perpendicular, and more and more is parallel