## ksaimouli Group Title 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? one year ago one year ago

1. henpen Group Title

|dw:1349633059599:dw|

2. ksaimouli Group Title

|dw:1349633142335:dw|

3. ksaimouli Group Title

what to do after this i am totally confused

4. ksaimouli Group Title

i know that ff=u*fn

5. ksaimouli Group Title

we need to find fn

6. henpen Group Title

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

7. ksaimouli Group Title

i got this but how do u find the F

8. henpen Group Title

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

9. ksaimouli Group Title

how did u get 117.6

10. henpen Group Title

mg

11. henpen Group Title

12*9.8

12. henpen Group Title

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

13. ksaimouli Group Title

thx

14. ksaimouli Group Title

why =fcos(25)

15. henpen Group Title

|dw:1349634909926:dw|Basic trig

16. henpen Group Title

$F_f=F_x=Fcos(25)$

17. henpen Group Title

No acceleration => no net force in x-direction