Question

A dynamometer D, is connected with two bodies of mass $M = 6kg$ and $m = 4kg$ . If two forces $F = 20N$ and $F = 10N$ are applied on masses according to figure then reading of the dynamometer will be:

A) $10N$
B) $20N$
C) $6N$
D) $14N$

Answer 1

Hint: In order to solve this question you have to know the concept of a dynamometer. As its name itself suggests ‘dyno’ which is the CGS unit of force, it is the device that is used for measuring the value of force, torque or moment of force, or power being applied to its ends. Also keep in mind that when there are two blocks attached and force is applied, then tension is experienced on both ends. The dynamometer reading gives the value of that tension.

Formula used:
$\sum {{F_x} = ma} $
Where $m$ is the mass of the object on which forces are applied
$a$ is the acceleration

Complete step by step solution:
Here in the question, the mass of the dynamometer is not given, so let us assume it to be very light or say massless. So the net force on the dynamometer is zero as the mass is zero. Hence the forces on both ends of the dynamometer must be equal and then the tension is the same on both ends.
Let us assume that the tension is T and the acceleration is a, now draw the free body diagram of both the blocks separately
Firstly, for the block of mass $M = 6kg$
Now, applying the formula that the net force is equal to the product of mass and acceleration,
$\sum {{F_x} = ma} $
Here the value of force is greater than the tension, so on putting values we have
$ \Rightarrow 20 - T = 6a$ ……………(i)
Now, draw the FBD of the block having mass $m = 4kg$
Again apply the formula, we have
$\sum {{F_x} = ma} $
Here, the motion of the block is towards left direction so the tension is greater than the force, so on putting the values we have
$ \Rightarrow T - 10 = 4a$ ……………(ii)
Now, from equation (ii), we get
$ \Rightarrow a = \dfrac{{T - 10}}{4}$
On putting this value of acceleration in equation (i), we get
$ \Rightarrow 20 - T = 6 \times \left( {\dfrac{{T - 10}}{4}} \right)$
\[ \Rightarrow 20 - T = \dfrac{{3T}}{2} - 15\]
On further solving that, we have
$ \Rightarrow \dfrac{{5T}}{2} = 35$
$\therefore T = 14N$
Hence the reading of the dynamometer is $14N$

Therefore, the correct option is (D).

Note: When there is no mass of dynamometer is given in the question then consider it as a massless but sometimes the mass is given then consider it and then solve the numerical. Also remember to consider the direction of the forces, as it changes the answer entirely.