two people are pulling a boat through water. Each exerts a force of 600N directed at a 30.0 degree angle relative to the forward motion of the boat. If the boat moves with constant velocity, find the resistive force (F) exerted by the water on the boat
2007-06-05 05:01:03 · 4 answers · asked by Anonymous in Science & Mathematics ➔ Physics
Let's assume that the forward motion is in the +x direction
The horizontal components of the two forces sum.
F_x = 600Cos30 + 600 Cos30 = 1039 N
F_y = 600Sin30 - 600Sin30 = 0
If the boat moves with constant velocity, then the net force acting on the boat is zero. hence the forward force of 1039 N must be balanced by the resistive force.
2007-06-05 05:06:59 · answer #1 · answered by dudara 4 · 0⤊ 0⤋
Each is exerting a force F along a line which is at 30 degrees to the direction of the boat's movement. So, the component of force along the direction of the boat's movement is Fcos30 and since there are two of them, the total force is 2Fcos30 = 1200.sqrt3/2 = 1039. N and since the boat is moving with a constant velocity, the acceleration is zero and hence the frictional force of water (viscosity) with the boat is equal and opposite to this.
2007-06-05 12:24:26 · answer #2 · answered by Swamy 7 · 0⤊ 0⤋
Each person exerts a force of 600*cos(30) = 519.6 N on the boat in the direction of motion. So the total force exerted in the direction of motion is 1,039.2 N and, since the boat is moving with constant speed, the net force on it must be 0 so the drag from the water is 1,039.2 N.
Doug
2007-06-05 12:10:14 · answer #3 · answered by doug_donaghue 7 · 0⤊ 0⤋
Each person exerts a FORWARD force of F * cos (angle from forward). The sideways forces cancel.
So the water's force opposing them is 2*F*cos (angle).
2007-06-05 12:05:34 · answer #4 · answered by Anonymous · 1⤊ 1⤋