The sum of the magnitudes of two forces is 25 | vedclass

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(A) Let the two forces be $F_1$ and $F_2$, where $F_1$ is the smaller force.Given: $F_1 + F_2 = 25$ (Equation $1$)The resultant $R$ is perpendicular t

Vedclass · Accepted Answer

$14.5 \,N, 10.5 \,N$

Vedclass · Answer

(A) Let the two forces be $F_1$ and $F_2$, where $F_1$ is the smaller force.Given: $F_1 + F_2 = 25$ (Equation $1$)The resultant $R$ is perpendicular to $F_1$. The magnitude of the resultant is $R = 10 \,N$.Using the triangle law of vector addition, since $R \perp F_1$, we have a right-angled triangle formed by $F_1$, $R$, and $F_2$ (as the hypotenuse).By Pythagoras theorem: $F_2^2 = F_1^2 + R^2$$F_2^2 - F_1^2 = 10^2 = 100$$(F_2 - F_1)(F_2 + F_1) = 100$Substitute $F_1 + F_2 = 25$ into the equation:$(F_2 - F_1)(25) = 100$$F_2 - F_1 = 4$ (Equation $2$)Adding Equation $1$ and Equation $2$:$2F_2 = 29 \implies F_2 = 14.5 \,N$Subtracting Equation $2$ from Equation $1$:$2F_1 = 21 \implies F_1 = 10.5 \,N$Thus, the two forces are $14.5 \,N$ and $10.5 \,N$.

The sum of the magnitudes of two forces is $25 \,N$. The resultant of these forces is perpendicular to the smaller force and has a magnitude of $10 \,N$. The two forces are:

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