Question

The sum of the magnitudes of two forces is $18 \mathrm{~N}$ and the magnitude of their resultant is $12 \mathrm{~N}.If the resultant makes an angle of $90^{\circ}$ with the smaller force, then find the magnitude of the forces.

$10 \mathrm{~N}, 8 \mathrm{~N}$

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Answer

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Given the sum of the magnitudes of two forces is $18 , \text{N}$ and the magnitude of their resultant is $12 , \text{N}$. The resultant makes an angle of $90^{\circ}$ with the smaller force. We need to find the magnitudes of the forces.

Let the magnitudes of the two forces be $P$ and $Q$ where $P < Q$.

From the problem statement: $$P + Q = 18 \qquad \Rightarrow \qquad P = 18 - Q \quad \ldots \ldots (1)$$

Given: $$R = 12 , \text{N}=\sqrt{P^{2} + Q^{2} + 2PQ \cos \theta}$$

Since the resultant makes an angle $\phi = 90^{\circ}$ with the smaller force ($P$), we use: $$\tan \phi = \frac{Q \sin \theta}{P + Q \cos \theta}$$
For $\phi = 90^{\circ}$: $$\Rightarrow \frac{Q \sin \theta}{P + Q \cos \theta} = \tan 90^{\circ} = \infty$$

Which implies: $$P + Q \cos \theta = 0 \quad \ldots \ldots \text{ (2)}$$

From Equation (1) and (2): $$Q (1 - \cos \theta) = 18 \quad \ldots \ldots \text{ (3)}$$

Rewriting the resultant force equation: $$R^2 = P^2 + Q^2 + 2PQ \cos \theta$$ Thus, $$12^2 = P^2 + Q^2 + 2PQ \cos \theta$$ Substitute $P = 18 - Q$ from Equation (1): $$144 = (18 - Q)^2 + Q^2 + 2(18 - Q)Q \cos \theta$$

Additionally, from Equations (1) and (2): $$R^2 = (P + Q)^2 - 2PQ (1 - \cos \theta)$$ $$\Rightarrow (18)^2 - 144 = 2PQ(1 - \cos \theta)$$ $$\Rightarrow 324 - 144 = 2PQ(1 - \cos \theta)$$ $$\Rightarrow 180 = 2PQ(1 - \cos \theta)$$ $$\Rightarrow PQ(1 - \cos \theta) = 90 \quad \ldots \ldots \text{ (4)}$$

Dividing Equation (4) by Equation (3): $$\frac{PQ(1 - \cos \theta)}{Q(1 - \cos \theta)} = \frac{90}{18}$$ $$\Rightarrow P = 5 , \text{N}$$

Using $P + Q = 18$ from Equation (1): $$5 + Q = 18$$ $$\Rightarrow Q = 13 , \text{N}$$

Thus, the magnitudes of the forces are $5 , \text{N}$ and $13 , \text{N}$.


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