The magnitude of the two forces forming a couple is 36 N each and the arm of the couple is 4 m . The magnitude of each force of an equivalent couple whose arm is 9 m , is (in Newtons)

Question

The magnitude of the two forces forming a couple is 36 N each and the arm of the couple is 4 m . The magnitude of each force of an equivalent couple whose arm is 9 m , is (in Newtons) (A) 18 (B) 26 (C) 16 (D) 15

Tardigrade Admin · Accepted Answer

Moment of couple = vecF × vec A B where F is force and A B is distance. Since, F1=36 N , d1=4 m , F2=F( say ) and d2=9 m Also, F1 × d1=F2 × d2 ⇒ 36 × 4=F × 9 ⇒ =16 N

Q. The magnitude of the two forces forming a couple is $36 N$ each and the arm of the couple is $4 m$ . The magnitude of each force of an equivalent couple whose arm is $9 m$ , is (in Newtons)

18

26

16

15

Moment of couple $= F \times A B$
where $F$ is force and $A B$ is distance.
Since, $F_{1} = 36 N, d_{1} = 4 m, F_{2} = F ($ say $)$
and $d_{2} = 9 m$
Also, $F_{1} \times d_{1} = F_{2} \times d_{2}$
$\Rightarrow 36 \times 4 = F \times 9$
$\Rightarrow= 16 N$

Q. The magnitude of the two forces forming a couple is 36N each and the arm of the couple is 4m . The magnitude of each force of an equivalent couple whose arm is 9m , is (in Newtons)

18

26

16

15

Moment of couple =F×AB where F is force and AB is distance. Since, F1​=36N,d1​=4m,F2​=F( say ) and d2​=9m Also, F1​×d1​=F2​×d2​ ⇒36×4=F×9 ⇒=16N

Q. The magnitude of the two forces forming a couple is $36 N$ each and the arm of the couple is $4 m$ . The magnitude of each force of an equivalent couple whose arm is $9 m$ , is (in Newtons)

Moment of couple $= F \times A B$
where $F$ is force and $A B$ is distance.
Since, $F_{1} = 36 N, d_{1} = 4 m, F_{2} = F ($ say $)$
and $d_{2} = 9 m$
Also, $F_{1} \times d_{1} = F_{2} \times d_{2}$
$\Rightarrow 36 \times 4 = F \times 9$
$\Rightarrow= 16 N$