# Thread: Need help getting started..

1. ## Need help getting started..

Sled of 196 N is being pulled across a horizontal surface at a constant velocity. So it's in equilibrium right??
The pulling force(Tension force?) has a magnitude of 80 n and is direct at an angle 30degree above the horizontal. Determine the coefficient of the kinetic friction.

I am stuck on drawing the FBD. I suck at those. I get them confused.
Need help on getting started, thanks.

So 30 degree angle from the 0degree, Force of 80 is perpendicular to that, so at 120 degrees?? Where do I put the Ff? to figure out the u

Sled of 196 N is being pulled across a horizontal surface at a constant velocity. So it's in equilibrium right??
The pulling force(Tension force?) has a magnitude of 80 n and is direct at an angle 30degree above the horizontal. Determine the coefficient of the kinetic friction.

I am stuck on drawing the FBD. I suck at those. I get them confused.
Need help on getting started, thanks.

So 30 degree angle from the 0degree, Force of 80 is perpendicular to that, so at 120 degrees?? Where do I put the Ff? to figure out the u
Here is the free body diagram

$\displaystyle \displaystyle \sum F_x=80\cos(30^{\circ})-\mu_k\eta=ma_x=0$

$\displaystyle \displaystyle \sum F_y=\eta+80\sin(30^{\circ})-196=ma_y=0$

3. Ok, so do I solve for Fy and sub that into Fx? What letter do you give the 80N just P for pull or T for tension?
I solve for Fn for Fy and plug that into Fn in Fx?

Thanks for the diagram I was off somewhat for sure.

4. Solve for eta$\displaystyle (\eta)$ the normal force in the 2nd equation.

Sub this into the first equation and solve for the coefficient of kinetic friction $\displaystyle \mu_k$

5. OK that is what I meant, I typed it wrong. Thank you. I will study that.