Solve the following :

Question:

The friction coefficient between the table and the blockshown in figure is $0.2$. Find the tensions in the two strings.

Solution:

$15 g-T=15 a \quad-(i)$

$T-T_{1}-\mu N=5 a \& \mu N=0.2 \times 5 g=g$

$T-T_{1}-g=5 a-(i i)$

$T_{1}-5 g=5 a \quad-(i i i)$

Adding (i), (ii) and (iii)

$15 g-g-5 g=15 a+5 a+50$

$9 q=25 a$

$a=\frac{9 g}{25}$

Now, put in eq.(i)

$T=15 g-15\left(\frac{9 g}{25}\right)$

$\mathrm{T}=96 \mathrm{~N}$

From eq.(3)

$T_{1}=5 g+5 \times \frac{9 g}{25}$

$T_{1}=68 \mathrm{~N}$

 

 

Leave a comment

Close

Click here to get exam-ready with eSaral

For making your preparation journey smoother of JEE, NEET and Class 8 to 10, grab our app now.

Download Now