A force vector F of magnitude 12N is applied to a FedEx box of mass m2=1kg. The force is directed up a plane tilted by theta=37 degrees. The box is connected by a cord to a UPS box of mass m1=3kg on the floor. The floor, plane, and pulley are frictionless, and the masses of the pulley and cord are negligible. What is the tension in the cord?

Note: the FedEx box is position on the 37 degree ramp and the UPS box is level on the ground. 12N is parallel to m2 and is pulling at the end of it.
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m2
___m1___(pulley)/ ramp at 37 degrees

It's too hard. You should give up now

The picture I get from your description is of mass m1 stationary on the ground while m2 gsts pushed up the ramp. This would create slack in the cord that connects the masses, passing over a pulley, and the tension would be zero.

This is probably not the picture you have in mind.

Agree with Charlie, but try finding the net force and use Newton's 2nd law maybe.

To determine the tension in the cord, we need to analyze the forces acting on the system.

Let's start by considering the forces acting on the FedEx box (m2):
1. Weight force (mg2): The weight of the FedEx box acts vertically downward, which can be calculated as mg2 = 1 kg * 9.8 m/s^2 = 9.8 N.
2. Tension force (T): The tension in the cord acts parallel to the inclined plane and in the opposite direction of the weight force.

Next, let's consider the forces acting on the UPS box (m1):
1. Weight force (mg1): The weight of the UPS box acts vertically downward, which can be calculated as mg1 = 3 kg * 9.8 m/s^2 = 29.4 N.
2. Tension force (T): The tension in the cord acts vertically upward, connecting the UPS box to the FedEx box through the pulley.

Since the floor, plane, and pulley are frictionless, there are no frictional forces to consider.

Now, based on the setup, the tension force (T) acting on both boxes is the same. This is because the cord connecting them is assumed to be massless and inextensible, so the tension remains constant throughout its length.

To find the tension in the cord, we need to analyze the forces acting on the FedEx box (m2) using Newton's second law in the vertical direction:
ΣFy(m2) = T - mg2 * sin(θ) = m2 * a(m2) = 0
Here, ΣFy(m2) represents the net force acting on the FedEx box in the vertical direction, sin(θ) represents the sine of the angle of the inclined plane (37 degrees), a(m2) represents the acceleration of the FedEx box (which is zero since it is not moving vertically), and T represents the tension force.

Simplifying the equation, we have:
T - mg2 * sin(θ) = 0
T = mg2 * sin(θ)
T = 1 kg * 9.8 m/s^2 * sin(37°)
T ≈ 5.882 N

Therefore, the tension in the cord is approximately 5.882 N.