One end of uniform rod of mass M and cross sectional area A is suspended from rigid support and an equal mass M is suspended at the other end . What is the stress at the mid point?

Open in App

Solution

Weight of suspended mass:

$W_{1} = m g$

Weight of the rod acting at the mid point:

Only half of the weight acts at the midpoint since the rod is uniform.

$W_{2} = \frac{m g}{2}$

Stress at the midpoint:

$S t r e s s = \frac{W_{1} + W_{2}}{A} = \frac{m g + \frac{m g}{2}}{A} = \frac{3 m g}{2 A}$

Suggest Corrections

Similar questions

One end of a uniform rod of mass M and cross-sectional area A is suspended from a rigid support and an equal mass M is suspended from the other end. The stress at the mid-point of the rod will be

Q. One end of a uniform rod of mass

M

and cross-sectional area

A

is suspended from the other end. The stress at the mid-point of the rod will be :

Q. One end of a uniform bar of weight

w_{1}

is suspended from the roof and a weight

w_{2}

is suspended from the other end, the area of cross-section is A. What is the stress at the mid point of the rod?

Join BYJU'S Learning Program

One end of uniform rod of mass M and cross sectional area A is suspended from rigid support and an equal mass M is suspended at the other end . What is the stress at the mid point?

Weight of suspended mass: W1=mg Weight of the rod acting at the mid point: Only half of the weight acts at the midpoint since the rod is uniform. W2=mg2 Stress at the midpoint: Stress=W1+W2A =mg+mg2A =3mg2A

Weight of suspended mass:

$W_{1} = m g$

Weight of the rod acting at the mid point:

Only half of the weight acts at the midpoint since the rod is uniform.

$W_{2} = \frac{m g}{2}$

Stress at the midpoint:

$S t r e s s = \frac{W_{1} + W_{2}}{A} = \frac{m g + \frac{m g}{2}}{A} = \frac{3 m g}{2 A}$