17a. A body of mass 5 kg is placed on a smooth... | WAEC SSCE - Further Mathematics - 2025 (Essay)

Question 1 Report

17a. A body of mass 5 kg is placed on a smooth plane inclined at an angle of 30º to the horizontal. Find: the magnitude of the force acting parallel to the plane.

bi. A uniform plank PQ of length 10m and mass m kg rests on two support A and B. Where \PA\ = \BQ\ = 1m. A load of mass 8kg is placed on the plank at point C such that \AC\ = 3.5m, if the reaction at B is 100N. Calculate the value of m

bii. the reaction at A [ take g = 10m/s\(^2\)].

Answer

17a. Since the plane is smooth (frictionless), the only force acting parallel to the inclined plane is the component of the weight of the body down the plane.

F = mg sin 30\(^\circ\)
Given m = 5 kg and taking g = 10 m/s\(^2\).
F = 5 \(\times 10 \times \sin 30^\circ\) = 5 \(\times 10 \times\) 0.5 = 25 N.
The magnitude of the force acting parallel to the plane is 25 N.

bi. The uniform plank PQ is 10 m long with supports at A and B such that PA = 1 m and BQ = 1 m. Thus, AB = 8 m. The plank’s centre of mass G is at its midpoint (5 m from P), so AG = 4 m. The 8 kg load is placed at C, where AC = 3.5 m.

Weight of plank = \( 10m \) N
Weight of load = \( 8 \times 10 = 80 \) N
Reaction at B, \( R_B = 100 \) N (given).

For rotational equilibrium, take moments about A (clockwise = anticlockwise):
\((10m) \times 4 + 80 \times 3.5 = 100 \times 8\)
40m + 280 = 800
40m = 520 \(\implies m = \frac{520}{40}\) = 13
Thus, m = 13 kg.

bii. For vertical equilibrium: \(R_A + R_B = \text{total weight}\)
\(R_A + 100 = (10 \times\) 13) + 80
R\(_A\) + 100 = 130 + 80 = 210
R\(_A\) = 210 - 100 = 110 N.
The reaction at A is 110 N.

Answer Details

17a. Since the plane is smooth (frictionless), the only force acting parallel to the inclined plane is the component of the weight of the body down the plane.

Weight of plank = \( 10m \) N
Weight of load = \( 8 \times 10 = 80 \) N
Reaction at B, \( R_B = 100 \) N (given).

bii. For vertical equilibrium: \(R_A + R_B = \text{total weight}\)
\(R_A + 100 = (10 \times\) 13) + 80
R\(_A\) + 100 = 130 + 80 = 210
R\(_A\) = 210 - 100 = 110 N.
The reaction at A is 110 N.

View Answer

17a. A body of mass 5 kg is placed on a smooth plane inclined at an angle of 30º to the horizontal. Find: the magnitude of the force acting parallel to the ...

Download The App On Google Playstore

17a. A body of mass 5 kg is placed on a smooth plane inclined at an angle of 30º to the horizontal. Find: the magnitude of the force acting parallel to the ...

Related Questions

Download The App On Google Playstore