Four forces acting at a point are in equilibrium. Three of the forces are 10 N at $0^\circ$, 15 N at $60^\circ$, and 20 N at $150^\circ$ (angles measured counterclockwise from the positive x-axis). What is the magnitude and direction of the fourth force?

Two forces $\overrightarrow {{{\rm{F}}_1}} \;\,and\,\overrightarrow {{{\rm{F}}_2}}$ are acting at right angles to each other. Then their resultant is

The velocity of a body at time t=0 is $10\sqrt 2 \;m/s$ in the north-east direction and it is moving with an acceleration of $2\;m/{s^2}$ directed towards the south. The magnitude and direction of the velocity of the body after 5 sec will be

A particle is subjected to three concurrent forces. Two of the forces have magnitudes of 5 N and 7 N and act at an angle of 60° to each other. If the particle is in equilibrium, what is the minimum possible magnitude of the third force?

The velocity of a body at time t=0 is $10\sqrt 2 \;m/s$ in the north-east direction and it is moving with an acceleration of $2\;m/{s^2}$ directed towards the south. The magnitude and direction of the velocity of the body after 5 sec will be

Two forces of 5 N and 10 N are acting at a point with an angle of 60° between them. Which of the following cannot be the magnitude of the resultant force?

The simple sum of two co-initial vectors is 10 units. Their vector sum is 8 units. The resultant of the vectors is perpendicular to the smaller vector. The magnitudes of the two vectors are

A body is under the action of two mutually perpendicular forces of 3 N and 4 N. The resultant force acting on the body is

If the magnitude of the sum of the two vectors is equal to the difference of their magnitudes, then the angle between vectors is

The resultant of two forces, each P, acting at an angle $heta$ is

Two forces of 3 N and 4 N act at a point at right angles to each other. The magnitude of their resultant is: