Q. 34

# Find the accelera

Given: 5a + T − 5g = 0

From free-body diagram (1), T = 5g − 5a …..(i)

T 8g 16a = 0

From free-body diagram (2),

T = 8g + 16a ……(ii)

From equations (i) and (ii), we get:

5g − 5a = 8g + 16a

So, the acceleration of the 5 kg mass is 97 m/s2 upward and that of the 4 kg mass is 2a= downward.

(b) From free body diagram (3),

8a T = 0

T = 8a

Again, T + 5a − 5g = 0

From free body diagram-4, 8a + 5a − 5g = 0

13a 5g = 0

downward

(c) T + 1a − 1g = 0

From free body diagram, T = 1g − 1a …..(i)

Again, from free body diagram

T 4g 8a = 0 ..(ii)

From equation (i) 1g − 1a − 4g − 8a = 0

downward

Acceleration of mass 1 kg is upward, Acceleration of mass 2 kg is downward

Rate this question :

How useful is this solution?
We strive to provide quality solutions. Please rate us to serve you better.
Try our Mini CourseMaster Important Topics in 7 DaysLearn from IITians, NITians, Doctors & Academic Experts
Dedicated counsellor for each student
24X7 Doubt Resolution
Daily Report Card
Detailed Performance Evaluation
view all courses
RELATED QUESTIONS :

A rectangular boxPhysics - Exemplar

A helicopter of mPhysics - Exemplar

There are four foPhysics - Exemplar

When a body <spanPhysics - Exemplar

Block <span lang=Physics - Exemplar

A block <span lanPhysics - Exemplar

Two <span lang="EPhysics - Exemplar