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

Question

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 (A) 7 N (B) 1 N (C) 5 N (D) zero

Tardigrade Admin · Accepted Answer

Resultant force, F=√F12+F22+2 F1 F2 cos θ =√(3)2+(4)2+2 × 3 × 4 cos 90° =5 N

Q. 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

7 N

1 N

5 N

zero

Resultant force, $F = F_{1}^{2} + F_{2}^{2} + 2 F_{1} F_{2} cos θ$
$= (3)^{2} + (4)^{2} + 2 \times 3 \times 4 cos 9 0^{\circ}$
$= 5 N$

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

7 N

1 N

5 N

zero

Resultant force, F=F12​+F22​+2F1​F2​cosθ​ =(3)2+(4)2+2×3×4cos90∘​ =5N

Q. 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

Resultant force, $F = F_{1}^{2} + F_{2}^{2} + 2 F_{1} F_{2} cos θ$
$= (3)^{2} + (4)^{2} + 2 \times 3 \times 4 cos 9 0^{\circ}$
$= 5 N$