Water Skier Problem

[mattx118]

A 92 kg water skier is being pulled at a constant velocity. The horizontal pulling force is 380 N.

The question is asking for what is the total resistive force exerted on the skier by the water and air.

I am stumped at the moment how to calculate that if anyone can help.

[Doc Al]

A 92 kg water skier is being pulled at a constant velocity.
What does that statement allow you to conclude about the net force on the skier?

[Tide]

The keywords are "constant velocity" meaning there is no acceleration! What does that tell you about the forces acting on the skier?

[mattx118]

That means acceleration = zero right? But that doesnt help me because in F=ma, if a = 0 then (0)(92) = 0 for Force, and thats not correct.

[mattx118]

I need to figure out the amoutn of friction thats what the problem is asking for i think

[Doc Al]

That means acceleration = zero right? But that doesnt help me because in F=ma, if a = 0 then (0)(92) = 0 for Force, and thats not correct.
Actually it is correct, but you have to know what it means. The F in F = ma stands for the net force on the object. Just because the net force is zero, doesn't mean no forces are acting---it just means they add to zero.

Identify the (horizontal) forces that act on this skier.

[mattx118]

Well it says theres 380 N horizontal pulling force, is that 380 net force? Meaning including Frictional Force and The applied force?

[Doc Al]

That 380 N is not the net (meaning total) force, it's only one force. (We already know the net force is zero, right?)

Here's a big hint: If there's a force of 380 N pulling one way, what must be the magnitude and direction of the resistive force (the only other force) if they must balance out---add to zero?

When it finally clicks you'll think that it couldn't be that easy. But it is. :smile:

[mattx118]

Or would that mean if the skier is moving at constant velocity that they two forces are at equilibrium and the frictional force would be 380 also or -380?

[Doc Al]

Or would that mean if the skier is moving at constant velocity that they two forces are at equilibrium and the frictional force would be 380 also or -380?
Absolutely!

[mattx118]

Ahh that is very easy I got the answer now. Thanks a lot

[mattx118]

Well I'm having trouble with a harder one than that if your up for helping :). I know I probably have a minor problem with it. Heres the question.

A 292-kg log is pulled up a ramp by means of a rope that is parallel to the surface of the ramp. The ramp is inclined at 34.0° with respect to the horizontal. The coefficient of kinetic friction between the log and the ramp is 0.880, and the log has an acceleration of 0.700 m/s2. Find the tension in the rope.

Ok heres what I got. I got myself a diagram drawn and I have the mg force exerted on the log going down which is (292*9.8) = 2714.6 which gives me my right side of my triangle. And the angle of incline is at 34 and I put the that angle in the top portion of my triangle because the angles are similar. So i've done Tan(22) * 2714.6 to find the bottom portion of the traingle or F(parallel) which I got as 1096.77. So that is the downward force acting on the Log. I also need to find the Frictional Force, which i got as 2518.21 (which is from the equation Fk=MkFn). And so those together are the forces acted down on the log. So my final equation i came up with is
( Ft-(Fk-Fparallel) ) / 292 = 0.700 .... ( Ft - 4448.98 ) / 292 = 0.700 and I got Ft to be 4653.38. However, thats not right. Is there something I did wrong, and I hope you can understand my work its hard typing it out here.

[Doc Al]

One problem I see is that you messed up in finding the parallel and perpendicular components of the log's weight. Rethink those triangles: The weight (mg) is the hypotenuse of the triangle, the components are the other two sides. The component of the weight parallel to the incline is mg sin\theta, perpendicular to the incline (equal to the normal force) is mg cos\theta.