The sum of the magnitudes of two forces actin | vedclass

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Question

(A) Let the two forces be $F_1$ and $F_2$, where $F_1$ is the smaller force. Let $F_1 = x$.Given that the sum of magnitudes is $16 \,N$, so $F_2 = 16

Vedclass · Accepted Answer

$6 \,N, 10 \,N$

Vedclass · Answer

(A) Let the two forces be $F_1$ and $F_2$, where $F_1$ is the smaller force. Let $F_1 = x$.Given that the sum of magnitudes is $16 \,N$, so $F_2 = 16 - x$.The resultant force $R = 8 \,N$ is normal to the smaller force $F_1$.Using the vector triangle or the property of the resultant, we have the relation: $F_2^2 = F_1^2 + R^2$.Substituting the values: $(16 - x)^2 = x^2 + 8^2$.Expanding the equation: $256 + x^2 - 32x = x^2 + 64$.Subtracting $x^2$ from both sides: $256 - 32x = 64$.Rearranging to solve for $x$: $32x = 256 - 64 = 192$.$x = \frac{192}{32} = 6 \,N$.Therefore, the smaller force $F_1 = 6 \,N$ and the larger force $F_2 = 16 - 6 = 10 \,N$.The forces are $6 \,N$ and $10 \,N$.

The sum of the magnitudes of two forces acting at a point is $16 \,N$. If their resultant is normal to the smaller force and has a magnitude of $8 \,N$, then the forces are:

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