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`        Two blocks 2kg & 4kg are connected through a massless string. Coeff. of friction btw 2kg & ground is 0.4 & coeff. of friction btw 4kg & ground is 0.6. F1=10N is applied on 2kg block to the left and F2=20N is applied on 4kg block to the right. Calculate frictional force btw 4kg block and ground. (assume tension is zero before F1 and F2 is applied).Pls solve it.`
7 years ago

147 Points
```										Dear Aditya
maximum friction force that can act in between 2 kg block and ground is =0.2*20 = 4
maximum friction force that can act in between 4 kg block and ground is  =0.6*40 = 24

when 2kg block just start moving toward left
equation on 2 kg block
so 10 -T- 4 =0
T =6
equation on 4 kg block
calculate 20-T = 14
which is less than the max friction force for 4 kg block so it will remain in the state of rest
and frictional force is 14 .
Please feel free to post as many doubts on our discussion forum as you can.If you find any question Difficult to understand - post it here and we will get you the answer and detailed  solution very  quickly. We are all IITians and here to help you in your IIT JEE preparation.All the best. Regards,Askiitians ExpertsBadiuddin

```
7 years ago
8 Points
```										i got the answer as 18 N. Briefly explain it.
```
7 years ago
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