A horse pulls a barge down a canal. The horse pulls on the rope with a force of 7900 N at an angle of 18 degrees to the direction of motion of the barge, which is heading straigt along the canal. The mass of the barge is 9500kg and its acceleration is .12m/s^2. What are the a) magnitude and b) direction of the force on the barge from the water?
Would I use F=mg sin (theda)? How would I find the direction?
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To find the magnitude of the force on the barge from the water, you would indeed use the equation F = mg sin(θ), where F is the force, m is the mass of the barge, g is the acceleration due to gravity (approximately 9.8 m/s^2), and θ is the angle between the force and the upward vertical direction.
In this case, the force acting on the barge is the horizontal component of the force exerted by the horse. The force exerted by the horse can be calculated by using the equation F = ma, where F is the force, m is the mass of the horse plus barge system, and a is the acceleration of the system.
First, calculate the force exerted by the horse:
F = ma
F = (m_horse + m_barge) * a
F = (mass of the horse + mass of the barge) * acceleration
F = (m_h + m_b) * a
F = (m_barge + m_barge) * a
F = 2m_barge * a
Substitute the given values into the equation:
F = 2 * (9500 kg) * (0.12 m/s^2)
F = 2280 N
Now, find the magnitude of the force on the barge from the water:
F_water = F * sin(θ)
F_water = 2280 N * sin(18 degrees)
F_water ≈ 645.56 N
So, the magnitude of the force on the barge from the water is approximately 645.56 N.
To find the direction of the force, you need to determine the angle between the force and the positive x-axis (direction of motion). Since the force is horizontal and acting opposite to the motion, the direction would be 180 degrees (opposite to the angle 0 degrees, which represents the positive x-axis).
Therefore, the direction of the force on the barge from the water is 180 degrees.