When the rocket launches it produces a downward force of 30000N. Due to the third law of motion, the rocket experiences a 30000N reaction force that pushes it upwards. 

In addition, the rocket experiences the downward force of its own weight. This is given by:

\(\displaystyle W = mg = 1000 kg * 10 \frac{m}{s^2} = 10000 N\)

We know that \(\displaystyle F_{net} = ma\).

We know the mass of the rocket is 1000kg, so we need only to find the net force to solve for acceleration.

We know that \(\displaystyle F_{net} = F - W\) (since \(\displaystyle F\) is directed upwards and \(\displaystyle W\) is directed downwards).

 \(\displaystyle F_{net} = 30000 N - 10000 N = 20000 N\)

Finally we solve for acceleration:

\(\displaystyle a = \frac{F_{net}}{m}=\frac{20000N}{1000kg}=20\frac{m}{s^2}\)

Since the boxes are all connected by ropes, we know that the acceleration of each box is exactly the same. They all move simultaneously, in tandem, with the same velocity and acceleration.

A quick analysis of each box will produce a very simple system of equations. Each rope experiences some tension, so we will label the tension experienced by the rope between \(\displaystyle m\) and \(\displaystyle 2m\) as \(\displaystyle T_{1}\). Similarly, we will label the tension experienced by the rope between \(\displaystyle 3m\) and \(\displaystyle m\) as \(\displaystyle T_{2}\).

For the box on the right (of mass \(\displaystyle 2m\)) we know that the force \(\displaystyle F\) pulls to the right and the rope pulls it to the left with a tension of \(\displaystyle T_{1}\).

We get the equation: \(\displaystyle F - T_{1} = 2ma\). \(\displaystyle 2m\) is the mass of the box and \(\displaystyle a\) is its acceleration (which is the same for the other boxes).

For the middle box (of mass \(\displaystyle m\)) we have a rope pulling on it on the right with tension \(\displaystyle T_{1}\) and a rope pulling on the left with tension \(\displaystyle T_{2}\).

So we get the equation: \(\displaystyle T_{1}- T_{2}= ma\). \(\displaystyle m\) is the mass of the box and \(\displaystyle a\) is its acceleration (which is the same for the other boxes).

Finally, for the box on the left (of mass \(\displaystyle 3m\)) we have only one rope pulling it to the right with a tension \(\displaystyle T_{2}\).

So we get the equation: \(\displaystyle T_{2}=3ma\). \(\displaystyle 3m\) is the mass of the box and \(\displaystyle a\) is its acceleration (which is the same for the other boxes).

Now it is just a matter of simple substitution. We have three equations:

\(\displaystyle T_{2}=3ma\)

\(\displaystyle T_{1}- T_{2}= ma\)

\(\displaystyle F - T_{1} = 2ma\)

From them we can get an expression for the force. Isolate \(\displaystyle T_{1}\) in the second equation.

\(\displaystyle T_{1}= ma + T_{2}\)

Substitute the expression of \(\displaystyle T_{2}\) from the first equation and simplify.

\(\displaystyle T_1= ma + 3ma = 4ma\)

Use this value of \(\displaystyle T_{1}\) in the third equation to get an expression for the force.

\(\displaystyle F = 2ma + T_{1} = 2ma + 4ma = 6ma\)

So we have:

\(\displaystyle F = 6ma\)

Solve for acceleration.

\(\displaystyle a = \frac{F}{6m}\)

This question requires a 2-dimensional analysis. First identify all the forces acting on the box. Since the box is being pushed against a surface it automatically experiences a normal force \(\displaystyle N\). The box experiences the downward force of its own weight given by \(\displaystyle mg\). Finally, since the box is trying to slide down, friction \(\displaystyle f\) opposes this motion. The following diagram shows these forces:

The key is that the box should NOT move.

For the horizontal axis, net force must be zero. The two horizontal forces are the applied force and the normal force.

\(\displaystyle F - N = 0N\)

\(\displaystyle F = N\)

Now, on the vertical axis we have fiction and the object's weight:

\(\displaystyle f - mg = 0N\)

\(\displaystyle f = mg\)

Finally, we use the equation for friction force to solve the problem.

\(\displaystyle f= N\mu\)

Substitute the weight, since it is equal to the force of friction.

\(\displaystyle N\mu=mg\)

Isolate the normal force.

\(\displaystyle N=\frac{mg}{\mu }\)

Since the normal force is equal to the applied force, this is our final expression.

\(\displaystyle F = N = \frac{mg}{\mu }\)

You need only to obtain the horizontal component of the force. To do this you must use trigonometric properties.

We see that the force makes an angle of 60º with the horizontal axis. This means that the horizontal component of the force is adjacent to this angle. We can view the diagram as a triangle. From trigonometry we know that to calculate the adjacent side of a triangle we need to multiply the hypotenuse by the cosine of the angle.

\(\displaystyle \cos\theta=\frac{adj}{hyp}\rightarrow adj=(hyp)(\cos\theta)\)

We can solve for the horizontal component of the force using the applied force as the hypotenuse:

\(\displaystyle F_{x} = Fcos(60º) = 20N*\frac{1}{2} = 10N\)