(a) Forces \(F_{1} = (3N, 210°)\) and \(F_{2} =... | WAEC SSCE - Further Mathematics - 2011 (Essay)

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(a) Forces \(F_{1} = (3N, 210°)\) and \(F_{2} = (4N, 120°)\) act on a particle of mass 7kg which is at rest. Calculate the :

(i) acceleration of the particle ; (ii) velocity of the particle after 3 seconds.

(b) \(F_{1} = (2i + 3j)N, F_{2} = (-5j)N \) and \(F_{3} = (6i - 4j)N\) act on a body. Find the magnitude and direction of the fourth force that will keep the body in equilibrium.

Answer

(a) Resolve the polar forces into components.

\[F_1=(3,210^\circ):\ (3\cos210^\circ,3\sin210^\circ)=(-2.598,-1.5).\] \[F_2=(4,120^\circ):\ (4\cos120^\circ,4\sin120^\circ)=(-2,3.464).\]

Resultant \(R=(-4.598,\ 1.964)\), so \(|R|=\sqrt{(-4.598)^2+(1.964)^2}=\sqrt{25.0}=5\ \text{N}\), at direction \(\theta=180^\circ-\tan^{-1}\!\frac{1.964}{4.598}=156.9^\circ\).

(i) Acceleration: \(a=\dfrac{|R|}{m}=\dfrac{5}{7}=0.71\ \text{m/s}^2\) in the direction \(156.9^\circ\).

(ii) Velocity after 3 s (from rest): \(v=at=0.71\times 3=2.14\ \text{m/s}\), along \(156.9^\circ\).

(b) Sum of the three forces: \(F_1+F_2+F_3=(2+0+6,\ 3-5-4)=(8,-6)\). The fourth force for equilibrium is \(F_4=-(8,-6)=(-8,6)\).

\[|F_4|=\sqrt{(-8)^2+6^2}=\sqrt{100}=10\ \text{N},\quad \theta=180^\circ-\tan^{-1}\!\tfrac{6}{8}=143.1^\circ.\]

The fourth force is 10 N at \(143.1^\circ\) to the positive x-axis.

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Answer Details

(a) Resolve the polar forces into components.

\[F_1=(3,210^\circ):\ (3\cos210^\circ,3\sin210^\circ)=(-2.598,-1.5).\] \[F_2=(4,120^\circ):\ (4\cos120^\circ,4\sin120^\circ)=(-2,3.464).\]

Resultant \(R=(-4.598,\ 1.964)\), so \(|R|=\sqrt{(-4.598)^2+(1.964)^2}=\sqrt{25.0}=5\ \text{N}\), at direction \(\theta=180^\circ-\tan^{-1}\!\frac{1.964}{4.598}=156.9^\circ\).

(i) Acceleration: \(a=\dfrac{|R|}{m}=\dfrac{5}{7}=0.71\ \text{m/s}^2\) in the direction \(156.9^\circ\).

(ii) Velocity after 3 s (from rest): \(v=at=0.71\times 3=2.14\ \text{m/s}\), along \(156.9^\circ\).

(b) Sum of the three forces: \(F_1+F_2+F_3=(2+0+6,\ 3-5-4)=(8,-6)\). The fourth force for equilibrium is \(F_4=-(8,-6)=(-8,6)\).

\[|F_4|=\sqrt{(-8)^2+6^2}=\sqrt{100}=10\ \text{N},\quad \theta=180^\circ-\tan^{-1}\!\tfrac{6}{8}=143.1^\circ.\]

The fourth force is 10 N at \(143.1^\circ\) to the positive x-axis.

Read lesson note on Vectors (WAEC)

Vectors

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(a) Forces \(F_{1} = (3N, 210°)\) and \(F_{2} = (4N, 120°)\) act on a particle of mass 7kg which is at rest. Calculate the : (i) acceleration of the particl...

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(a) Forces \(F_{1} = (3N, 210°)\) and \(F_{2} = (4N, 120°)\) act on a particle of mass 7kg which is at rest. Calculate the : (i) acceleration of the particl...

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