## ksaimouli 3 years ago A block is pulled by a string that makes an angle of 25º to the horizontal. If the mass of the block is 12.0 kg and the coefficient of friction is 0.25, what force would keep the block moving at a constant velocity?

1. anonymous

|dw:1349633059599:dw|

2. ksaimouli

|dw:1349633142335:dw|

3. ksaimouli

what to do after this i am totally confused

4. ksaimouli

i know that ff=u*fn

5. ksaimouli

we need to find fn

6. anonymous

$F_n=mg-F_y=12*9.8-Fsin(25)$

7. ksaimouli

i got this but how do u find the F

8. anonymous

$F_f= \mu F_n= 0.25(117.6-Fsin(25))=F_x=Fcos(25)$

9. ksaimouli

how did u get 117.6

10. anonymous

mg

11. anonymous

12*9.8

12. anonymous

$29.4-0.25\sin(25)F=Fcos(25)$ $\frac{29.4}{\cos(25)+0.25\sin(25)}=F$

13. ksaimouli

thx

14. ksaimouli

why =fcos(25)

15. anonymous

|dw:1349634909926:dw|Basic trig

16. anonymous

$F_f=F_x=Fcos(25)$

17. anonymous

No acceleration => no net force in x-direction