A student gets his car stuck in a snow drift. Not at a loss, having studied physics, he attaches one end of a stout rope to the vehicle and the other end to the trunk of a nearby tree, allowing for a small amount of slack. The student then exerts a force vector F on the center of the rope in the direction perpendicular to the car-tree line, as shown in the figure below. If the rope is inextensible and if the magnitude of the applied force is 489 N, what is the force on the car? (Assume equilibrium conditions.)
Well, I guess this student really knows how to "pull" off a physics trick! The force exerted on the car is equal to the force exerted by the car, but in the opposite direction. So, if the student is applying a force of 489 N perpendicular to the car-tree line, the force on the car will also be 489 N but in the opposite direction. It's like a tug of war with physics!
To solve this problem, we can use the concept of vector decomposition. Let's break down the given force vector F into its components.
The given force vector F is acting perpendicular to the car-tree line, which means it can be decomposed into two components: one parallel to the car-tree line and the other perpendicular to it.
The perpendicular component of the force does not affect the car since it is perpendicular to the car's direction. Thus, it does not contribute to the force on the car.
The parallel component of the force is responsible for exerting a force on the car in the direction of the car-tree line. This is the force that will help free the car from the snow drift.
Therefore, to find the force on the car, we need to determine the magnitude of the parallel component of the force.
Let's denote the magnitude of the force on the car as F_c and the angle between the force vector F and the car-tree line as θ.
Since the force vector F is perpendicular to the car-tree line, the angle θ is 90 degrees.
The parallel component of the force can be found using the equation:
F_c = F * cos(θ)
Since cos(90) is zero, the force on the car in this case would be:
F_c = 489 N * 0 = 0 N
Therefore, the force on the car is zero.
To find the force on the car, we need to analyze the forces acting on the system. In this case, the forces involved are the force applied by the student (F) and the tension in the rope.
Since the system is in equilibrium (no net force or acceleration), we know that the magnitude of the force applied by the student is equal to the magnitude of the tension in the rope.
To solve this problem, we can use the concept of vector components. Let's say the angle between the force applied by the student and the car-tree line is θ.
In this case, the force applied by the student can be decomposed into two components: one in the direction of the rope (F_parallel) and one perpendicular to the rope (F_perpendicular).
F_parallel = F * cos(θ)
F_perpendicular = F * sin(θ)
Since the system is in equilibrium, the magnitude of F_parallel should be equal to the magnitude of the tension in the rope.
Therefore, the force on the car is equal to F_parallel.
Now, we need to determine the value of F_parallel. Since we know the magnitude of the applied force (F = 489 N), we need to find the angle θ.
Unfortunately, the figure mentioned in your question is not provided, so we don't have a specific angle to work with. However, assuming θ is given or you can determine it from additional information, you can substitute it into the equation to find F_parallel.
Keep in mind, this calculation assumes an idealized situation without taking into account other factors such as friction or the weight of the car. In reality, the force required to move the car would depend on these additional factors.
Therefore, to find the force on the car, you need to know the angle θ between the force applied by the student and the car-tree line. Once you have that information, substitute it into the equation F_parallel = F * cos(θ) to calculate the force on the car.
Force on the car in the direction of the rope is
F = (489/2)÷sin(θ)
where θ=angle formed by the rope and the car-tree line.