Recent content by yoshi6

Can anyone tell me if this is right?

Ok this is what I have come up with::: F(friction) = 524cos12= 513N so now I have, F(friction) =(uk)(Fn) Fn= F(friction)/ (uk) = 513/0.15 =3420 Weight= 3420- 524sin12 =3308 So, Weight = mg m=weight/g =3308/9.81 =340kg ? I hope...

oh! F(normal force) = mg = 524-109

I think so...524-109=mg?

I don't know. I don't understand. Thanks anyway.

I do have a free body diagram that my teacher said is correct. I know normal force =mg. But if I don't have the mass how can I find the normal force? I know I have to substract the frictional force but again I need the mass to figure this out. Right?

So that would be the Fy value? Which is 524Sin12= 108.945726 then I add this value to the Fx value? So, F(normal force) = 108.945726+512.5493428 =621.5 N ?

but the question is asking for the weight...is it not?

Can someone please help me

Homework Statement A crate on a floor has an applied force (12 degrees below horizontal) of 524N. The crate moves with constant velocity. The coefficient of kinetic friction is 0.15. What is the mass of the crate? Homework Equations Fk=uk*Fn, Fn= m * g The Attempt at a Solution...

can someone please be so kind and check this for me? Thanks!

okay so here is my attempt... convert---> 1.01600 to meters and 86 km squared to meters and then... 0.01016 x 9270 x 9270 = 873, 078.3m^2 now to compare 24 hours in seconds/ 5.8m^2 per second= 86400/ 33.64=2568.37 So, therefore, the volume of water over Guelph city is greater than...

no nevermind, i don't think it is working out properly...hmm

THANKYOU! I see now

pleaseeeee help!