Physics: maximum force that can be applied to move an object?

What is the maximum force F which can be exerted?
Given: The friction between the block with
mass 14 kg and the wedge with mass 16 kg
is 0.19 . The surface between the wedge with
mass 16 kg and the horizontal plane is smooth
(without friction).
The acceleration of gravity is 9.8 m/s2 .
A block is released on the inclined plane 21 degrees with the horizontal
(top side of the wedge).
What is the maximum force F which can
be exerted on the 16 kg block in order that
the 14 kg block does not move up the plane?
Answer in units of N.

This is what I did:
Vertical force - weight parallel - friction = 0

Fsin21 - 14*9.8 sin21 - 0.19(14*9.8 cos21) = 0
Fsin21 - 137.2 sin21 - 0.19(137.2 cos21) = 0
0.3584F - 49.1681 - 24.3366 = 0
0.3584F = 73.5047
F = 205.09 N

I typed in this answer and it was incorrect. Did I do something wrong in the computation process?

Your computation process seems correct up to a point, but there is a mistake in the last step. Let me walk you through the calculation again to identify the error.

First, let's identify the forces acting on the 14 kg block on the inclined plane. These forces are:

1. Weight of the block acting vertically downwards: m1 * g * cos(21°)
2. Component of the weight acting parallel to the plane: m1 * g * sin(21°)
3. Friction force opposing the motion: μ * N, where N is the normal force (N = m1 * g * cos(21°))

Since the block is not moving up the plane, the force exerted, F, should be equal to or less than the sum of these forces.

So, the equation becomes:

F ≤ m1 * g * sin(21°) + μ * (m1 * g * cos(21°))

Plugging in the given values:
m1 = 14 kg
μ = 0.19
g = 9.8 m/s^2

F ≤ 14 * 9.8 * sin(21°) + 0.19 * (14 * 9.8 * cos(21°))

Now we can calculate this expression:

F ≤ 65.804 + 26.151

F ≤ 91.955 N

Therefore, the maximum force F that can be exerted on the 16 kg block in order for the 14 kg block not to move up the plane is approximately 91.955 N.