Question
Question: A uniform horizontal meter scale of mass m is suspended by two vertical strings attached to its two ...
A uniform horizontal meter scale of mass m is suspended by two vertical strings attached to its two ends. A body of mass 2m is placed on the 75cm mark. The tensions in the two strings are in the ratio:
(a) 1:2
(b) 1:3
(c) 2:3
(d) 3:4
Solution
Hint So here to solve this type of problem, we should have the basic knowledge of the tension. Firstly we will take out the rotational equilibrium which will be corresponding to the T1. And in the same way, we will find them T2 , and then we will be able to find the ratio of the tension in the string.
Complete Step By Step Solution First of all we all know, according to the question the net tension produced will be equal to the sum of the two tension applied to it.
Here,
⇒T1+T2=2mg+1mg
So it will be equal to the
⇒T1+T2=3mg
Now taking tension about the left side of the plank, we will get
⇒0.5mg+0.75×2mg=1×T2
And from here, on solving the above equation we will the values as
⇒T2=2mg
And similarly, another torque will be
⇒T1=mg
Now we have to find the ratios, therefore the ratio of the tension will be given by
⇒T2T1=2mgmg
And on solving the above, we will get
⇒T2T1=21
Therefore, we can say that
21, will be the ratio of the tension in the strings.
Hence, the option (a) is correct.
Note As we know the pressure force is the total force compressed matter exerts longitudinally - i.e. the sum of pressure x cross-section area across all parts of the total cross-section. (In math terms, the integral of pressure across the total cross-section.)
So the tension is just the opposite: The integral of tensile stress across the total cross-section. In other words, the total force that tries to stretch a body of matter.
So we can say that the pressure or, respectively, tensile stress (the difference is only in the sign) in the matter is commonly called stress.