Two boxes with different masses M1 = 1.8 kg and M2 = 2.6 kg are tied together on a frictionless ramp surface which makes an angle θ = 21° with the horizontal (see the figure below).

What is the tension in the rope connecting the two boxes? N

What is the tension in the rope connecting the first box to the ramp? N

No figure or description was provided

There was a ramp the two box where sliding toward to left seems like a 45 degree angle n first is m1 thn m2 does this helppp

To determine the tension in the rope connecting the two boxes and the tension in the rope connecting the first box to the ramp, we can make use of Newton's second law and the principles of equilibrium.

1. Tension between the two boxes:
Since the ramp is frictionless, there are no horizontal forces acting on the boxes. Therefore, the only vertical forces are the weights of the boxes (M1*g for the first box, and M2*g for the second box) and the tension in the rope between them.
The vertical component of the weight of the first box is M1*g*sin(θ), and the vertical component of the weight of the second box is M2*g*sin(θ). These vertical components must be balanced by the tension in the rope.
Therefore, the tension in the rope connecting the two boxes can be calculated as:
Tension between the two boxes = M1*g*sin(θ) + M2*g*sin(θ)

2. Tension between the first box and the ramp:
To determine the tension in the rope connecting the first box to the ramp, we need to consider the horizontal forces acting on the first box. The horizontal component of the weight of the first box is M1*g*cos(θ). This force must be balanced by the tension in the rope between the first box and the ramp.
Therefore, the tension in the rope connecting the first box to the ramp can be calculated as:
Tension between the first box and the ramp = M1*g*cos(θ)

Substituting the given values of the masses (M1 = 1.8 kg, M2 = 2.6 kg), angle θ (21°), and gravitational acceleration (g ≈ 9.8 m/s^2), you can calculate the tensions in the ropes.