An adventurer pulls a loaded sled with a mass of 56kg along a horizontal meadow at a constant speed. The coefficent of kinetic force between the sled and the grasee is .45 and the angle of the rope between the sled and the adventurer is 55 degrees. What is the tension in the rope?

Question

anonymous · Answer

Hey I'll help!

anonymous · Answer

in x direction: 0 = Tension*cos55 - Fnormal*0.45
((mg)*(0.45))/(cos55) = Tension

anonymous · Answer

I can imagine two solutions, neither of which might be right, the first one might be right, since Tension in the y direction can equal zero....or you can equate the x and the y directions and come up with something like this: 

Tcos55-mg0.45=sin55T and then use basic algebra to solve for T

anonymous · Answer

Since the acceleration is both axes are given(a=0), 
$\Sigma F_y =F_g=F_N+Tsin 55 $
$\Sigma F_x=F_f =Tcos 55 $
Now, since $F_f=\mu F_N$, 
sub $\mu=.45, m=56kg, g=9.8ms^{-2}$
We get,
$548.8 =F_N + 0.819T $
$0.45(F_N)=0.574T$
Solve the stimultanous eqs to get T.