## 1018 one year ago physics problem (question's in the comments)

1. 1018

2. Michele_Laino

3. 1018

oh hi ms michele! thank you for your help again. haha, ok i will :)

4. 1018

@Michele_Laino hi are you still online? im having trouble with the problem. if you're not too busy maybe you can help me? thanks

5. Michele_Laino

6. 1018

oh, that's just short for miss. haha. sorry

7. anonymous

Let's look at block m2 first forces acting on it are |dw:1441042822826:dw|

8. anonymous

for equilibrium resultant force on block m2=0 this implies for vertical components $N=m_{2}g+F.\sin(34)$ We know that friction is $f=\mu.N$ $\therefore f=\mu(m_{2}g+F.\sin(34))$ For horizontal components we have $T+f=F.\cos(34)$ From this you can find T $T=F.\cos(34)-f=F.\cos(34)-\mu(m_{2}g+F.\sin(34))$

9. anonymous

But then why is the block 1 given for?Maybe I'm missing out something :p

10. 1018

11. anonymous

Do you have an answer so we can at least verify by plugging in values?

12. 1018

no i dont sorry. actually the problem changed already, i mean same problem but with different values because i tried to answer it and it was wrong haha so im just trying to see the analysis of it

13. anonymous

oh

14. 1018

but do you want to see the new? although it's literally the same problem only with different values

15. 1018

or i could just type the new values. but really the same question.

16. 1018

i have an idea but i dont know. haha. there 2 tensions present in the string, so maybe it's tension (right) - tension (left) lol

17. anonymous

One thing that I think is not given is whether the applied force is keeping the system in equilibrium or causing acceleration, which is a key deciding factor Let's assume it causes acceleration a, cuz that seems to be the case here, Therefore for block 1 we have Force acting towards the right on it is the tension T and towards left have friction equal to normal reaction $T-\mu.m_{1}g=m_{1}a$ Thus to find T we need to eliminate a $T=m_{1}(\mu.g+a)$ Now consider the system of box 1 and 2 combined We have $a=\frac{F_{res}}{m_{1}+m_{2}}$ Where F is the resultant force on the system of 2 blocks I think F is $F_{res}=T-\mu.m_{1}g+F.\cos(34)-\mu(m_{2}g+F.\sin(34))-T$ The first 2 terms correspond to block 1 and other terms to block 2 cancelling out the Tension we get $F_{res}=F.\cos(34)-\mu(m_{1}g+m_{2}g+F.\sin(34))$ $\implies a=\frac{F.\cos(34)-\mu(m_{1}g+m_{2}g+F.\sin(34))}{m_{1}+m_{2}}$ plugging this into equation for Tension we have $T=m_{1}(\mu.g+\frac{F.\cos(34)-\mu(m_{1}g+m_{2}g+F.\sin(34))}{m_{1}+m_{2}})$ I'm not sure about it, but maybe worth a shot

18. 1018

yeah that's what i thought the first time, but i just assumed that it was not moving since there's a string involved.

19. anonymous

Can you try the expression I got for Tension?see if it works

20. 1018

ill try this equation. thanks! i'll let you know if its correct. haha

21. 1018

the dot is multiplication right? just to be clear, with the mew(?) and g

22. anonymous

yeep, I had to do it otherwise it was writing it as mug lol

23. anonymous

$mug$

24. 1018

on calcu its negative, is that ok?

25. 1018

oh hold on i think i made a mistake ill try again

26. 1018

hey! it's correct! thank you so much!