A boy pulls a toy having two carriages placed on a frictionless table by applying a force of 1.5 N at 30°in a vertical plane . Calculate (i)the acceleration of the 40g carriage. (ii)the acceleration of the 20g carriage. (iii)the tension in the string fastened between the two carriages. Show in the figure the forces acting on both separately.
Answers
Given: A boy pulls a toy having two carriages placed on a friction less table by applying a force of 1.5 N at 30°in a vertical plane .
To find: Calculate (i) the acceleration of the 40 g carriage.
(ii) the acceleration of the 20 g carriage.
(iii) the tension in string fastened between two carriages.
Solution:
- Now the components of Force are:
Vertical: 1.5 sin 30 = 1.5/2 = 7.5 N
Horizontal : 1.5 cos 30 = 1.5 x √3/2 = 0.75√3 N
- Acceleration for 20 gm will be:
Total mass will be: 0.04 kg + 0.02 kg = 0.06 kg
a = F/m = 0.75√3 N/ 0.06 kg
a = 21.65 m/s²
- Now acceleration for 40 gm will be:
A = √(0.75√3/ 0.06)² + (0.75 - 0.4/ 0.04)²
After solving, we get:
A = 23.34 m/s²
- Then the Tension will be:
T = ma
T = 0.02 x 21.65
T = 0.43 N
Answer:
So the acceleration of the 40 g carriage is 23.34 m/s², acceleration of the 20 g carriage is 21.65 m/s² and the tension in string fastened between two carriages is 0.43 N.
I. 21.65 m/s²
23.34 m/ s²
0.43N
