Search any question & find its solution
Question:
Answered & Verified by Expert
Consider an object kept at the centre, in the $X Y$-plane, on which five coplanar forces act as shown in the figure. The resultant force on the object is
Options:
Solution:
1526 Upvotes
Verified Answer
The correct answer is:
$6.5 \mathrm{~N}, 330^{\circ}$
The total force in $X$-direction,
$\begin{aligned}
& \Sigma F_x=15 \cos 60^{\circ}+19-11 \cos 30^{\circ}-16 \cos 45^{\circ} \\
& \Sigma F_x=5.66 \mathrm{~N}
\end{aligned}$
The total force in $Y$-direction,
$\begin{aligned}
& \Sigma F_y=-22+15 \sin 60^{\circ}+16 \cos 45^{\circ}-11 \sin 30^{\circ} \\
& \Sigma F_y=-3 \cdot 196 \mathrm{~N}
\end{aligned}$
Then, resultant force
$\begin{aligned}
\Rightarrow \quad R & =\sqrt{F_x^2+F_y^2}=\sqrt{(5 \cdot 66)^2+(3 \cdot 196)^2} \\
R & =6 \cdot 5 \mathrm{~N}
\end{aligned}$
Direction of resultant,
$\begin{aligned}
\Rightarrow & \tan \theta & =F_y / F_x=\frac{-3 \cdot 196}{5 \cdot 66} \\
\Rightarrow & \theta & =330^{\circ}
\end{aligned}$
Looking for more such questions to practice?
Download the MARKS App - The ultimate prep app for IIT JEE & NEET with chapter-wise PYQs, revision notes, formula sheets, custom tests & much more.