A farm tractor tows a 3800-kg trailer up a 19 degrees incline with a steady speed of 3.0 m/s. What force does the tractor exert on the trailer? (Ignore friction.)
draw FBD's
force of tractor up19 degrees
=force of gravity along that axis
To find the force exerted by the tractor on the trailer, we need to analyze the forces acting on the trailer on the inclined plane.
There are two main forces at play: the force exerted by the tractor, and the force of gravity acting on the trailer. Since the trailer is moving with a steady speed, we can assume that the net force acting on it is zero.
First, let's find the force of gravity acting on the trailer. The force of gravity can be calculated using the equation:
F_gravity = m * g
where F_gravity is the force of gravity, m is the mass of the trailer, and g is the acceleration due to gravity. In this case, the mass of the trailer is given as 3800 kg and the acceleration due to gravity can be taken as 9.8 m/s^2.
F_gravity = 3800 kg * 9.8 m/s^2 = 37240 N
Next, we need to resolve the force of gravity into its components parallel and perpendicular to the inclined plane. The component of the force of gravity acting parallel to the incline is:
F_parallel = F_gravity * sin(θ)
where θ is the angle of incline (19 degrees in this case).
F_parallel = 37240 N * sin(19°) = 12123 N
Since the trailer is moving at a steady speed, the force exerted by the tractor must be equal in magnitude and opposite in direction to the force of gravity acting parallel to the incline.
Therefore, the force exerted by the tractor on the trailer is:
F_tractor = -12123 N
So, the tractor exerts a force of -12123 N (opposite to the direction of motion) on the trailer.