# The sum of the magnitudes of two forces is 18 N and the magnitude of their resultant is 12 N. If the resultant makes an angle of 90 degrees with the smaller force, then find the magnitude of the forces. 10 N, 8 N

## 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

**:**

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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