Question

A body of mass m is suspended by two strings making angles alpha and beta with the horizontal find the tension on the string

Answer 1

Answer:

hey

Explanation:

T1cos a = t2 cos b

t2 =t1 cos a / cos b..........(i)

V.C. T1 and T2 balance mg

t1 sin a +t2 sin b =mg

from (i) ti sin a +ti cos a / cos b * sin b =mg

after solving .....

t1 mg cos b/ sin a cos b + cis a sin b

= mg cos b / sin (a+ b )

t2 = mg cos a / sin (a+ b)

its clear

Answer 2

Answer:

The tensions in the strings are $T_{1}=\frac{m g \cos \beta}{\sin (\alpha+\beta)}$ and $T_{2}=\frac{m g \cos \alpha}{\sin (\alpha+\beta)}$.

Explanation:

Firstly we will consider a graph illustrating the directions of the horizontal and vertical elements of the tension working on the string.

The graph is shown below,

The horizontal components of $T_{1}$and $T_{2}$balance each other as the body is in equilibrium.

$&T_{1} \cos \alpha=T_{2} \cos \beta$

$&T_{2}=\frac{T_{1} \cos \alpha}{\cos \beta} \ldots \ldots(1)$

The vertical components of $T_{1}$and $T_{2}$ balance the weight of the body.

$T_{1} \sin \alpha+T_{2} \sin \beta=m g \ldots$ (2)

Substitute the equation (1) in equation (2).

$&T_{1} \sin \alpha+\frac{T_{1} \cos \alpha}{\cos \beta} \sin \beta=m g$

$&\Rightarrow T_{1}=\frac{m g \cos \beta}{\sin \alpha \cos \beta+\cos \alpha \sin \beta}$

Using the trigonometric function, continue further computation.

$T_{1}=\frac{m g \cos \beta}{\sin (\alpha+\beta)}$

The trigonometric function that we have utilizes to solve the above equation is provided as follows.

$\sin (a+b)=\sin a \cos b+\sin b \cos a$

Hence, the tension in the first string is,

$T_{1}=\frac{m g \cos \beta}{\sin (\alpha+\beta)}$

Consider the equation and substitute the expression of tension in the first string.

$&T_{2}=\frac{m g \cos \beta}{\sin (\alpha+\beta)} \frac{\cos \alpha}{\cos \beta}$

Therefore $T_{2}=\frac{m g \cos \alpha}{\sin (\alpha+\beta)}$

Thus, the tension in the second string is, $T_{2}=\frac{m g \cos \alpha}{\sin (\alpha+\beta)}$.

Therefore, the tensions in the string are:$T_{1}=\frac{m g \cos \beta}{\sin (\alpha+\beta)}$ and

$T_{2}=\frac{m g \cos \alpha}{\sin (\alpha+\beta)}$

therefore The tensions in the strings are $T_{1}=\frac{m g \cos \beta}{\sin (\alpha+\beta)}$ and $T_{2}=\frac{m g \cos \alpha}{\sin (\alpha+\beta)}$.

#SPJ2