The best way to think of a system in equilibrium is that it is displaying a constant velocity in both the horizontal and vertical directions. Both the skydiver and the tightrope walker have a constant velocity. The driver, however, is driving around a circular track. You must remember that velocity is a vector of magnitude and direction. Because the direction of the driver is changing, he is not driving at a constant velocity; thus, he is not in equilibrium, and is experiencing an acceleration (centripetal acceleration).

Keep in mind that the skydiver is experiencing both the downward force of gravity, and the upward force of air resistence on his deployed parachute. These forces cancel, producing a net force of zero, and allowing him to fall at a constant velocity in equilibrium.

The motion of the boat () needs to counteract the motion of the man (). As the man's foot pushes him forward, the boat pushes back (Newton's third law). Due to opposing forces, the boat will move backward as the man moves forward.

This scenario can also be solved by looking at momentum. Both the man and the boat start from rest, so in order for momentum to be conserved once the man gains positive velocity, the boat must gain negative velocity.

Since the only acceleration on the object is to the west, we can conclude that there is no vertical acceleration on the object. As a result, the northern force must be offset by the third force. The only way to offset the northern force is to have a force pointing south, in any direction. As such, the third force could be directly south, southwest, or southeast as long as the southern component can counteract the northern force.

The third force would only need to be equal to the northern force if it was pointing directly due south, and we can only confirm that the third force is not completely offsetting the western force. As a result, we can only conclude that the third force has some magnitude in the southern direction.

There are only two forces acting on the box: the force of gravity and the force of tension in the rope. Since the box is not in motion, we can assume that the system is in equilibrium and the net force is equal to zero.

Rearranging, we can see that gravity and the force of tension will be equal and opposite.

Calculate the force of gravity using Newton's second law.

Note that the force of gravity is negative, since it acts in the downward direction. Use this value to solve for the force of tension.

Given that the system is motionless and that the tensional force is constant throughout the rope, we know that the weights on the platform must balance the weight of the mass. Given that the mass weights 2kg and each weight is 0.5kg, we can determine the number of weights required by dividing 2kg/0.5kg = 4 weights. Note that we can neglect the weight of the pulley in this problem; while this is usually the case on the MCAT as well, be sure to double check.

Newton's first law indicates that an object in motion will "want" to remain in motion at constant velocity unless it is acted on by a net force. In this case, Newton's first law is the only reasonable answer since in space, it is assumed there is no frictional or gravitational forces acting on the shuttle. Newton's second law can be described using the equation . Note that his first law is just a special case of the second law, where . Snell's law describes the interaction between waves and different media. Kirchhoff's laws involve charge conservation within circuits. Pascale's principle indicates that pressure exerted within a fluid is exerted equally in all directions.

Since the two vectors are perpendicular, they form a right triangle with the vector sum. Therefore, we simply use the Pythagorean theorem to solve for the sum.

The two vector components' directions determine the vector sum's direction as well. Since the components are north and east, the resultant vector is roughly northeast.

Use Newton's second law, , to relate the forces and accelerations of the particles. Writing this out for the two charges gives the following equations.

The two charges exert equal-magnitude and opposite-direction forces on one another according to Newton's third law, so , or using the right side of the second law equations, .

Plugging in our given values allows us to solve for .

Since the block is moving to the right, we know that the frictional force is acting to the left. Also acting to the left is the gravitational force on the hanging block. Since it is a constant velocity of , we know that acceleration is zero and the downward force on the hanging block plus the frictional force is exactly equal to the force . Add the two forces.

For acceleration to be zero, the force needs to equal the sum of the frictional force on the block and the gravitational force on the hanging block.

To then accelerate the system, an additional force is needed.

This is the additional force required. Thus the total force required is: