Question:

Two bodies of masses 15 kg and 10 kg are connected with light string kept on a smooth surface. A horizontal force F=500 N is applied to a 15 kg as shown in the figure.Calculate the tension acting in the string.

Solution:

To identify the forces acting on each mass:

• Gravitational force on the 15 kg mass acting downward.
• Gravitational force on the 10 kg mass acting downward.
• Applied force of 500 N acting left on the 15 kg mass.
• Tension T in the string for the 15 kg mass is acting in the direction opposite to the appled force.
• Tension in the string for the 10 Kg mass is acting in the direction of motion.

Acceleration of all the blocks and the string:

$${Acceleration =\frac{net \ force}{net \ mass}}$$

$${Acceleration =\frac{500 \ N}{25 \ Kg}}$$

$${Acceleration =20 \ \frac{m}{s^2}}$$

Free body diagram of the 15 Kg block:

Equation using Newton’s second law:

For the 15 kg mass:

$${F-T=m_1 a}$$

$${500-T=15 a}$$

$${T=500-15(20)}$$

$${T =200 \ N}$$

Tension between the 15 kg and 10 kg block is 200 N

For the 10 Kg mass

$${T_1 =10(20)}$$

$${T_1=200 N}$$

So the tension in the string is 200 N

ALITER:

Tension in the string is

$${T=\frac{m_2}{m_1+m_2} \ F}$$

$${T=\frac{10}{15+10} \ (500)}$$

$${T=200 \ N}$$

PHYSICS QUIZ:

https://forms.gle/FiZsuCGyPCDN7kXz8