Question

Figure 6-E3
13. The friction coefficient between the table and the block
shown in figure (6-E4) is 0-2. Find the tensions in the
two strings.
5 kg
2
5 kg
15 kg
Figure 6-E4​

Answer 1

Explanation:

From the FBD of blocks (15 kg block assumed to move downwards) and writing Newton's 2nd law of motion ,F=Ma

for 15 kg block : 15g−T

1

=15a ...(1)

for 5 kg hanging block : T

2

−5g=5a ...(2)

for 5 kg block on the horizontal surface, we use Newton's second law in two direction (i.eX−axis and Y−axis)

Balancing force along Y-direction : N=5g, and using concept of friction f=μN, we get f=0.2×5×g=g

Now along X-direction : T

1

−T

2

−f=5a ....(3)

using value of f and adding eq 1, eq 2 and eq 3 we get

⇒ 15g−5g−f=(15+5+5)×a

⇒ 9g=25a

⇒ a=

25

9g

=

25

9×10

=

5

18

s

2

m

(since g=10)

now using eq 1 : T

1

=150−

5

15×18

=150−54=96N

using eq 2 : T

2

=50+

5

5×18

=68N

Hence values of T

1

=96N and T

2

=68N

solution