In the following figure the masses of the blocks A and B are same and each equal to m. The tensions in the strings O A and A B are T2 and T1, respectively. The system is in equilibrium with a constant horizontal force mg on B. Then T1 is

Question

In the following figure the masses of the blocks A and B are same and each equal to m. The tensions in the strings O A and A B are T2 and T1, respectively. The system is in equilibrium with a constant horizontal force mg on B. Then T1 is (A) m g (B) √2 m g (C) √3 m g (D) √5 m g

Tardigrade Admin · Accepted Answer

Considering free body diagram of block B Sigma Fx=0 ; T1 cos θ1=m g ...(i) Sigma Fy=0 ; T1 cos θ1=-m g ...(ii) By squaring and adding equations (i) (ii), we get T12( sin 2 θ1+ cos 2 θ1)=2(m g)2 ∴ T1=√2 m g

Q. In the following figure the masses of the blocks $A$ and $B$ are same and each equal to $m$ . The tensions in the strings $O A$ and $A B$ are $T_{2}$ and $T_{1}$ , respectively. The system is in equilibrium with a constant horizontal force mg on $B$ . Then $T_{1}$ is

$m g$

$2 m g$

$3 m g$

$5 m g$

Considering free body diagram of block $B$
$Σ F_{x} = 0; T_{1} cos θ_{1} = m g$ ...(i)
$Σ F_{y} = 0; T_{1} cos θ_{1} = - m g$ ...(ii)
By squaring and adding equations
(i) & (ii), we get
$T_{1}^{2} (sin^{2} θ_{1} + cos^{2} θ_{1}) = 2 (m g)^{2}$
$∴ T_{1} = 2 m g$

Q. In the following figure the masses of the blocks A and B are same and each equal to m. The tensions in the strings OA and AB are T2​ and T1​, respectively. The system is in equilibrium with a constant horizontal force mg on B. Then T1​ is

mg

2​mg

3​mg

5​mg