Question

A body of mass m is suspended by two strings making angles $\alpha$ and $\beta$ with the horizontal (as shown in the figure). Find the tensions in the strings?

• A. $T_2=\frac{gcos\beta }{sin(\alpha +\beta )};T_2=\frac{gcos\alpha }{sin(\alpha +\beta )}$
• B. $T_2=\frac{mgcos\alpha }{sin(\alpha +\beta )};T_2=\frac{mgcos\beta }{sin(\alpha +\beta )}$
• C. $T_1=\frac{mgcos\beta }{sin(\alpha +\beta )};T_2=\frac{mgcos\alpha }{sin(\alpha +\beta )}$
• D. $T_2=mgcot\alpha ;T_2=mgtan\beta$

Answer 1

The correct option is C

$T_1=\frac{mgcos\beta }{sin(\alpha +\beta )};T_2=\frac{mgcos\alpha }{sin(\alpha +\beta )}$

Take the body of mass in as the system. The forces acting on the system are

(i) Mg downwards (by the earth),

(ii) $T_1$ along the first string (by the first string) and

(iii) $T_2$ along the second string (by the second string).

These forces are shown in figure. As the body is in equilibrium, these forces must add to zero. Taking horizontal components,

$T_{1}cos\alpha -T_{2}cos\beta =0$

$T_{1}cos\alpha =T_{2}cos\beta$ ------------------(i)

or,

Taking vertical components,

$T_{1}sin\alpha +T_{2}sin\beta -mg=0$-------------(ii)

Eliminating $T_2$ from (i) and (ii),

$T_{1}sin\alpha +T_1\frac{cos\alpha }{cos\beta }sin\beta =mg$

or,

$T_1=\frac{mg}{sin\alpha +\frac{cos\alpha }{cos\beta }sin\beta }=\frac{mgcos\beta }{sin(\alpha +\beta )}$

From(i),

$T_2=\frac{mgcos\alpha }{sin(\alpha +\beta )}$