Solving Tension & Friction Problem: Get Answer from Magellanic

[Magellanic]

Hi, I am having some major difficulties with this problem:

A sledge loaded with bricks has a total mass of 18.5 kg and is pulled at constant speed by a rope inclined at 20.2° above the horizontal. The sledge moves a distance of 20.9 m on a horizontal surface. The coefficient of kinetic friction between the sledge and surface is 0.500.

What is the tension in the rope?


I do not know what I am doing wrong, because every answer I get is incorrect.

So far I can figure out that the sledge weighs 181.3N.
When calculating the Frictional Force (mu * N), I come up with 90.65
Because the sledge is moving at a constant speed, there is no acceleration, and therefore the Sum of the horizontal forces = 0.

That would mean, when drawing out a force diagram, the Horizontal F = 90.65, calculating the Hypotenuse (Tension in the rope) at 96.59.

But that's not the answer!
I realize that the Vertical force is present due to the angle of the rope, and would thus subtract from the initial weight of the box, which would change the Frictional Force, continuing on to change the vertical force again. It's just a big loop.

So where am I screwing up/What equation am i missing?


Thanks,
Magellanic

 

[Stevedye56]

mu=tanTheta=.368 I am pretty sure that this is the coefficient of static friction since they gave you the coefficient of kinetic fricition.

 

[Magellanic]

That doesn't make sense. Why do I need to know the static friction if the sled is in motion?

 

[andrevdh]

Using the information given by the problem I also get the same answer.

 

[radou]

...
So far I can figure out that the sledge weighs 181.3N.
When calculating the Frictional Force (mu * N), I come up with 90.65
Because the sledge is moving at a constant speed, there is no acceleration, and therefore the Sum of the horizontal forces = 0...

You got N wrong, as it seems to me. N = m*g*cos(20.2) = 170.32 [N]. If the surface was horizontal, then N would equal 181.3 [N].