The sum of the magnitudes of two forces actin | vedclass

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(A) Let the two forces be $F_1$ and $F_2$,where $F_1$ is the smaller force.Given: $F_1 + F_2 = 16 ~N$,so $F_2 = 16 - F_1$.Let the resultant force be $

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$6 ~N, 10 ~N$

Vedclass · Answer

(A) Let the two forces be $F_1$ and $F_2$,where $F_1$ is the smaller force.Given: $F_1 + F_2 = 16 ~N$,so $F_2 = 16 - F_1$.Let the resultant force be $R = 8 ~N$.Since the resultant is perpendicular to the smaller force $F_1$,we have the relation $R^2 + F_1^2 = F_2^2$.Substituting the values: $8^2 + F_1^2 = (16 - F_1)^2$.$64 + F_1^2 = 256 + F_1^2 - 32 F_1$.$32 F_1 = 256 - 64$.$32 F_1 = 192$.$F_1 = 6 ~N$.Then,$F_2 = 16 - 6 = 10 ~N$.Thus,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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