Resultant and Component

Find the resolved parts of the force whose inclination to one of the resolved parts is given alongside as $50 \mathrm{kg} \mathrm{wt}, 60°$.

Find the resolved parts of the force whose inclination to one of the resolved parts is given alongside as $16 \mathrm{N}, 30°$. 

The law of parallelogram of forces gives the resultant of 

Derive the expression for resultant magnitude and direction of two force acting at a point.

State the law of parallelogram of two forces.

A force of $50 \mathrm{N}$ is making an angle of $30°$with x-axis, Find the magnitude of the vertical component.

A force is inclined  at $60°$to the horizontal. If the rectangular component in horizontal direction is $50 \mathrm{N}.$ Find the magnitude of the force and its vertical components respectively. 

A child pulls a rope attached to a stone with a force of $60 \mathrm{N}.$ The rope makes an angle of $60°$ degrees to the ground. What is the effective force along the horizontal direction?

The X- component of a force of $50 \mathrm{N}$ is $30 \mathrm{N}$, then what will be the Y-component of the same applied force?

A force of $10 \mathrm{N}$ is resolved into the perpendicular components. If the first component makes $30°$ with the force the magnitudes of the components are:
 

Define balanced forces.

If the forces are unbalanced, the resultant force is zero.

The resultant of balanced forces is :
 

Let us have three forces as $50 \mathrm{N}$, $60 \mathrm{N}$, and $20 \mathrm{N}$. These are acting on a body simultaneously. The direction of $20 \mathrm{N}$ is opposite to the direction of the other two. Determine the resultant force.

A body of mass $\mathrm{M}$ is kept on a rough horizontal surface (friction coefficient $\mu$). A person is trying to pull the body by applying a horizontal force but the body is not moving. The force by the surface on the body is $\mathrm{F},$ where

A $7 \mathrm{kg}$ object is subjected to two forces ${\overrightarrow{F}}_1=\left(20\hat{\mathrm{i}}+30\hat{\mathrm{j}}\right)$ and ${\overrightarrow{F}}_2=\left(8\hat{\mathrm{i}}+5\hat{\mathrm{j}}\right) \mathrm{N}$. The magnitude of resulting acceleration in $\mathrm{m} {\mathrm{s}}^{-2}$ will be

Two vectors $A$ and $B$ have equal magnitudes. The magnitude of $\left(A+B\right)$ is $n$ times the magnitude of $\left(A-B\right)$. The angle between $A$ and $B$ is

Two forces $P$ and $Q$ of magnitude $2F$ and $3F$, respectively, are at an angle $\mathrm{\theta }$ with each other. If the force $Q$ is doubled, then their resultant also gets doubled. Then, the angle $\mathrm{\theta }$ is

Find the magnitude of the resultant of shown forces in $XY$ plane.

A ball of mass $0.2 \mathrm{kg}$ is thrown vertically upwards by applying a constant force by hand. If the hand moves $0.2 \mathrm{m}$ while applying the force and the ball goes upto $2 \mathrm{m}$ height further, find the magnitude of the force. Consider $g=10 \mathrm{m} {\mathrm{s}}^{-2}$.