Solveeit Logo

Question

Question: (a) Find acceleration. (b) Find tension in the string. ![](https://www.vedantu.com/que...

(a) Find acceleration.
(b) Find tension in the string.

Explanation

Solution

The force acting on a body creates acceleration. Now if the body attached to the string feels force, then, there originates a tension on the string. These tension pulls the body attached to the other end of the string and thus produces an acceleration to the whole system.

Step by step answer:
Formulae Used:
If a force FF acts on an object of mm and produces acceleration aa then you have the expression
F=maF = ma
Given:
For the figure (1):
The force FF acted upon is 500N500N.
The mass of the immediately attached body is M=20kgM = 20kg.
The mass of the body attached to the other end is m=10kgm = 10kg.

For the figure (2):
The force FF acted upon is 500N500N.
The mass of the immediately attached body is m=10kgm = 10kg.
The mass of the body attached to the other end is M=20kgM = 20kg.
To get: (a) The acceleration.
(b) The tension in the string.
Step 1:
Let the tension on the string be TT. Let the acceleration on the system is aa.
For the figure (1) you can equate the forces acting on the system.
500T=20a500 - T = 20a
T=10aT = 10a
Putting eq (3) in eq (2) you have
50010a=20a 30a=500 a=50030=16.67  500 - 10a = 20a \\\ \Rightarrow 30a = 500 \\\ \Rightarrow a = \dfrac{{500}}{{30}} = 16.67 \\\
a=16.67m/s2\therefore a = 16.67m/{s^2}
So, calculate the value of TT
T=16.67×10=166.7T = 16.67 \times 10 = 166.7
T=166.7N\therefore T = 166.7N
Step 2:
Let the tension on the string be TT. Let the acceleration on the system is aa.
For the figure (2) you can equate the forces acting on the system.
500T=10a500 - T = 10a
T=20aT = 20a
Putting eq (3) in eq (2) you have
50020a=10a 30a=500 a=50030=16.67  500 - 20a = 10a \\\ \Rightarrow 30a = 500 \\\ \Rightarrow a = \dfrac{{500}}{{30}} = 16.67 \\\
a=16.67m/s2\therefore a = 16.67m/{s^2}
So, calculate the value of TT
T=16.67×20=333.4T = 16.67 \times 20 = 333.4
T=333.4N\therefore T = 333.4N

Final Answer:
From figure (1),
a) The acceleration of the system is 16.67ms216.67m{s^{ - 2}}.
b) The tension on the string is 166.7N166.7N.
From figure (2),
a) The acceleration of the system is 16.67ms216.67m{s^{ - 2}}.
b) The tension on the string is 333.4N333.4N.

Note: The tension on the string plays a crucial role in the system. The end where the force is applied the tension on the string is generated. Similarly due to this tension a reaction is generated from the other end of the string. Here no friction is considered. So, you should take the acceleration of the whole system the same.