i) F(y) = mg
ii)F(x) = 520 N
iii)
F(y) =mg (upward); mg (downward)
F=520 N ( to the right)
F(fr) = 520 N (to the left)
Net force =0 => Steady speed => equilibrium
A water-skier is being pulled at a steady speed in a straight line. her mass plus the mass of the ski is 65 kg. the pull of the tow rope on her is 520 N.
a...i) What is the vertical component Y of the push of the water on the ski??
....ii) What is the horizontal component X of the push of the water on the ski.( ignore air resistance)
....iii) Component X and the 520 N towing force form a clockwise couple acting on the water skier. explain how can she remain in equilibrium as she is towed along?
1 answer
1 answer