Relevant equations:

Determine the clockwise torque caused by the bucket.

Write an expression for the counterclockwise torque caused by the rope.

Combine the torque of the rope and the torque of the bucket into the net torque equation.

Since the system has no angular acceleration, net torque must be zero, allowing us to solve for the force of the rope.

The correct answer is external force. External forces applied to an object will result in non-zero acceleration, causing the object to move. In contrast, internal forces contribute to the properties of the object and do not result in acceleration of the object.

Either positive or negative forces can result in the acceleration of an object. The sign on the force simply conveys information about its relative direction.

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:

We know that .

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  (since  is directed upwards and  is directed downwards).

Finally we solve for acceleration:

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  and  as . Similarly, we will label the tension experienced by the rope between  and  as .

For the box on the right (of mass ) we know that the force  pulls to the right and the rope pulls it to the left with a tension of .

We get the equation: .  is the mass of the box and  is its acceleration (which is the same for the other boxes).

For the middle box (of mass ) we have a rope pulling on it on the right with tension  and a rope pulling on the left with tension .

So we get the equation: .  is the mass of the box and  is its acceleration (which is the same for the other boxes).

Finally, for the box on the left (of mass ) we have only one rope pulling it to the right with a tension .

So we get the equation: .  is the mass of the box and  is its acceleration (which is the same for the other boxes).

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

From them we can get an expression for the force. Isolate  in the second equation.

Substitute the expression of  from the first equation and simplify.

Use this value of  in the third equation to get an expression for the force.

So we have:

Solve for acceleration.

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 . The box experiences the downward force of its own weight given by . Finally, since the box is trying to slide down, friction  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.

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

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

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

Isolate the normal force.

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

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.

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