There are four forces acting at a point $P$ produced by strings as shown in the figure. The system is at rest. Find the forces $F_1$ and $F_2$.

Vedclass pdf generator app on play store
Vedclass iOS app on app store
(N/A) Given forces acting at point $P$ are $1 \text{ N}$ at $45^{\circ}$ to the vertical,$2 \text{ N}$ at $45^{\circ}$ to the vertical,$F_2$ along the horizontal,and $F_1$ along the vertical downward.
For the system to be in equilibrium,the net force must be zero: $\sum \overrightarrow{F} = 0$.
Resolving forces along the horizontal ($x$-axis) and vertical ($y$-axis) directions:
$\sum F_x = 0 \implies F_2 + 1 \cos 45^{\circ} - 2 \cos 45^{\circ} = 0$
$F_2 + \frac{1}{\sqrt{2}} - \frac{2}{\sqrt{2}} = 0$
$F_2 - \frac{1}{\sqrt{2}} = 0 \implies F_2 = \frac{1}{\sqrt{2}} \text{ N} \approx 0.707 \text{ N}$.
$\sum F_y = 0 \implies 1 \sin 45^{\circ} + 2 \sin 45^{\circ} - F_1 = 0$
$F_1 = 3 \sin 45^{\circ} = 3 \times \frac{1}{\sqrt{2}} = \frac{3}{\sqrt{2}} \text{ N} \approx 2.121 \text{ N}$.

Explore More

Similar Questions

Three forces start acting simultaneously on a particle moving with velocity $\vec{v}.$ These forces are represented in magnitude and direction by the three sides of a triangle $ABC$ (as shown). The particle will now move with velocity

Write the condition for equilibrium when two forces act on a particle.

$A$ $1 \,kg$ mass is suspended from the ceiling by a rope. $A$ horizontal force $F$ is applied at the midpoint of the rope so that the upper part of the rope makes an angle of $45^{\circ}$ with respect to the vertical axis as shown in the figure. The magnitude of $F$ is:

Three concurrent forces of the same magnitude are in equilibrium. What is the angle between the forces? Also,name the triangle formed by the forces as sides.

The adjoining figure shows a force of $40\, N$ acting at $30^o$ to the horizontal on a body of mass $5\, kg$ resting on a smooth horizontal surface. Assuming that the acceleration due to gravity is $10\, ms^{-2}$,which of the following statements is (are) correct?
$[1]$ The horizontal component of the applied force is $20\sqrt{3}\, N$.
$[2]$ The weight of the $5\, kg$ mass acts vertically downwards.
$[3]$ The net vertical force acting on the body is $30\, N$ (upwards).

Difficult
View Solution

Vedclass Products

For Students

Vedclass Test Series

Mock tests in real JEE/NEET style with performance analysis. 5-day free trial.

Start Free Trial
For Teachers

Exam Paper Generator

Generate Set A/B/C/D exam papers from 7.5L+ questions in 2 minutes. 3 chapters free.

Try Free
For Institutes

Online Exam Module

Live online exams with unlimited students, 360° analytics & white-label branding.

See Demo