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JAMB Physics 2023 Paper

Question : 26

Total: 65

Three forces with magnitudes 16 N, 12 N and 21 N are shown in the diagram below. Determine the magnitude of their resultant force and angle with the x-axis

7.63N,61°
8.71N,61°
7.63N,29°
8.71N,29°

Validate

Solution:

Concept:

Resolve the forces into horizontal and vertical components, then find the resultant magnitude and direction.

Explanation:

Take right and upward as positive.

Net horizontal component:

F_x = 12\cos 32^\circ - 16\cos 26^\circ = -4.20 \, \text{N}

This acts to the left.

Net vertical component:

F_y = 12\sin 32^\circ + 16\sin 26^\circ - 21 = -7.63 \, \text{N}

This acts downward.

Magnitude of the resultant:

R = \sqrt{(-4.20)^2 + (-7.63)^2}

R = \sqrt{17.64 + 58.22} = \sqrt{75.86}

R = 8.71 \, \text{N}

Angle with the x-axis:

\tan \theta = \frac{7.63}{4.20} = 1.8095

\theta = \tan^{-1}(1.8095) = 61^\circ

Answer:

8.71 \, \text{N}, 61^\circ

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