Two forces of magnitudes 3 , N and 4 , N are | vedclass

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(B) Given,the two forces are $A = 3 \, N$ and $B = 4 \, N$ with an angle $	heta = 90^\circ$ between them.Since the forces are perpendicular,the resul

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$5$

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(B) Given,the two forces are $A = 3 \, N$ and $B = 4 \, N$ with an angle $	heta = 90^\circ$ between them.Since the forces are perpendicular,the resultant force $R$ is given by:$R = \sqrt{A^2 + B^2 + 2AB \cos 	heta}$$R = \sqrt{3^2 + 4^2 + 2(3)(4) \cos 90^\circ}$Since $\cos 90^\circ = 0$,we have:$R = \sqrt{3^2 + 4^2} = \sqrt{9 + 16} = \sqrt{25} = 5 \, N$The direction of the resultant force with respect to the $4 \, N$ force is given by:$	an \alpha = \frac{3}{4} \implies \alpha = 	an^{-1}(0.75) \approx 37^\circ$.

Two forces of magnitudes $3 \, N$ and $4 \, N$ are acting on an object. If the angle between them is $90^\circ$,then their resultant force is ... $N$.

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