Two people are pulling a boat through the water. Each exerts a force of 600 N directed at a 30.0 angle relative to the forward motion of the boat. If the boat moves with constant velocity, find the resistive force exerted by the water on the boat.

We have that for the Question "Two people are pulling a boat through the water. Each exerts a force of 600 N directed at a 30.0 angle relative to the forward motion of the boat. If the boat moves with constant velocity, find the resistive force exerted by the water on the boat." it can be said that the resistive force exerted by the water on the boat is

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Two people are pulling a boat through the water. Each exerts a force of 600 N directed at a 30.0 angle relative to the forward motion of the boat. If the boat moves with constant velocity, find the resistive force exerted by the water on the boat.

Generally the equation for the Force  is mathematically given as

F=1039N

Therefore

the resistive force exerted by the water on the boat is

F=1039N

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It sounds as though the two people are standing in front of the boat on opposite sides of it, so that they both make an angle of 30.0° with the axis of the boat, as in the attached free body diagram (ignoring the force of buoyancy and the weight of the boat).

By Newton's second law, the net vertical force is

∑ F = P₁ sin(60.0°) + P₂ sin(120.0°) - R = 0

where upward is positive and downward is negative, and the right side is 0 because the boat moves with constant velocity and thus zero acceleration.

We're told that P₁ = P₂ = 600 N, and we know sin(60°) = sin(120°), so the above reduces to

R = 2 P sin(60.0°) = 2 (600 N) sin(60.0°) ≈ 1040 N

can be answered using momentum.

momentum = p

p1= mass1 * velocity1

p1 = p2

therefore

p1 = 6kg * 4m/s

p1 = 24kgm/s

p2=m2 * v2

24 kgm/s = 7kg * v2

v2 = 24 kgm/s / 7kg

v2= 3.42 m/s