Forces of $5 \sqrt{2} N$ and $6 \sqrt{2}$ are acting on a mass of $1000 k g$ at an angle of $60^{o}$ to each other. Find the acceleration produced in the mass.

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Solution

We know that the resultant force $(F) = \sqrt{F_{1}^{2} + F_{2}^{2} + 2 F_{1} F_{2} cos θ}$
$\Rightarrow F$ $= \sqrt{(5 \sqrt{2})^{2} + (6 \sqrt{2})^{2} + 2 (5 \sqrt{2}) (6 \sqrt{2}) cos 60^{0}}$
$\Rightarrow F = 13.49 N$
$\Rightarrow a = \frac{F}{m} = \frac{13.49}{1000} = 0.01349 m / s^{2}$

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Force $5 \sqrt{2} N$ and $6 \sqrt{2} N$ are acting on a mass of $1000 k g$ at an angle of $60^{\circ}$ to each other. What is the acceleration of the mass after $10$ second from the instant the forces started acting?

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Forces of 5√2N and 6√2 are acting on a mass of 1000 kg at an angle of 60o to each other. Find the acceleration produced in the mass.

We know that the resultant force (F)=√F21+F22+2F1F2cosθ⇒F=√(5√2)2+(6√2)2+2(5√2)(6√2)cos600⇒F=13.49N⇒a=Fm=13.491000=0.01349m/s2

Forces of $5 \sqrt{2} N$ and $6 \sqrt{2}$ are acting on a mass of $1000 k g$ at an angle of $60^{o}$ to each other. Find the acceleration produced in the mass.

We know that the resultant force $(F) = \sqrt{F_{1}^{2} + F_{2}^{2} + 2 F_{1} F_{2} cos θ}$
$\Rightarrow F$ $= \sqrt{(5 \sqrt{2})^{2} + (6 \sqrt{2})^{2} + 2 (5 \sqrt{2}) (6 \sqrt{2}) cos 60^{0}}$
$\Rightarrow F = 13.49 N$
$\Rightarrow a = \frac{F}{m} = \frac{13.49}{1000} = 0.01349 m / s^{2}$