A 85 kg water skier is being pulled at a constant velocity. The horizontal pulling force is 310 N.
(a) Find the magnitude of the total resistive force exerted on the skier by the water and air.
__________N
(b) Find the magnitude of the upward force exerted on the skier by the water.
__________N
2007-09-23 03:52:57 · 5 answers · asked by Anonymous in Science & Mathematics ➔ Physics
Constant velocity implies the forces must add up to zero (ie, cancel one another), so:
a) The resistive forces must equal the pulling force.
b) The upward force on the skier must equal the weight (mg)
They give you the force and the mass. You know g. Easy cheesy.
If this wasn't completely obvious to you, draw a picture with all the forces that have to cancel. Then it should be quite clear. Newton's third law never really comes into this at all.
2007-09-23 03:57:44 · answer #1 · answered by Anonymous · 0⤊ 0⤋
Because the velocity is constant the total resistance must be 310N, other wise there would be acceleration ( or deceleration).
Because the skier doesn't sink or rise into the air, the upward force must equal the skiers weight.
2007-09-23 11:01:28 · answer #2 · answered by Anonymous · 0⤊ 0⤋
(a) the sum of the forces is zero (equal and opposite forces). Therefore since the horizontal pulling force is 310N, the magnitude of the total resistive force is 310N.
(b) the force downward by the skier is F=mg, where g=9.81
F = 85(9.81) = 834N. Since equal and opposite forces, the upward force on the skier is 834N
2007-09-23 11:05:45 · answer #3 · answered by theanswerman 3 · 0⤊ 0⤋
This is really a "2nd law" problem, not a "3rd law" problem.
> "...pulled at a constant velocity."
That's an important clue. Whenever anything moves at a constant velocity, it means (by Newton's 1st & 2nd laws) that all the forces acting on it, cancel each other out.
It also means that, separately, all the horizontal forces cancel each other out; and all the vertical forces cancel each other out.
Horizontal forces:
1) pull of rope. This is 310 N
2) resistance of air & water friction. Since it exactly cancels #1, it must be exactly as strong as #1.
Vertical forces:
3) gravity pulling down. This equals mg = (85kg)(9.8m/s²)
4) water pushing up. Since it exactly cancels #3, it must be exactly as strong as #3.
2007-09-23 11:04:15 · answer #4 · answered by RickB 7 · 0⤊ 0⤋
For every action there is an equal and opposite reaction, is that what your talking about?
2007-09-23 11:00:02 · answer #5 · answered by 4 · 0⤊ 0⤋