## MCAT Physics and Math Review

**Chapter 1: Kinematics and Dynamics**

### 1.7 Mechanical Equilibrium

So far we’ve been paying attention to kinematics and the special cases of linear and projectile motion. However, many times the MCAT will require us to eliminate acceleration, or otherwise maintain a system in equilibrium. To accomplish this, we must be familiar with analyzing forces, especially with free body diagrams, as well as with the special conditions for translational and rotational equilibrium. The study of forces and torques is called **dynamics**.

FREE BODY DIAGRAMS

While we all have an intuitive sense of forces (and their effects) in everyday life, students often struggle to represent them diagrammatically. Drawing **free body diagrams** takes some practice but will be a valuable tool on the MCAT. On Test Day, make sure to draw a free body diagram for any problem in which you must perform calculations on forces.

**MCAT EXPERTISE**

When dealing with dynamics questions, always draw a quick picture of what is happening in the problem; this will keep everything in its proper relative position and help prevent you from making simple mistakes.

**Example:**

Three people are pulling on ropes tied to a tire with forces of 100 N, 150 N, and 200 N as shown below. Find the magnitude and direction of the resultant force. (Note: sin 30° = 0.5, sin 60° = 0.866)

**Solution:**

First, draw a free body diagram that shows the forces acting on the tire. Its purpose is to identify and visualize the acting forces.

The resultant force is simply the sum of the forces. To find the resultant force vector, we need the sum of the force components, shown below.

The net component vectors are shown graphically below.

The magnitude can then be determined using the Pythagorean theorem:

The direction can be determined using the tangent function:

Note that this calculation of an inverse tangent is beyond the scope of Test Day math. The resultant vector is shown below.

TRANSLATIONAL EQUILIBRIUM

Translational motion occurs when forces cause an object to move without any rotation. The simplest pathways may be linear, such as when a child slides down a snowy hill on a sled, or parabolic, as in the case of a cannonball shot out of a cannon. Any problem regarding translational motion in the *Chemical and Physical Foundations of Biological Systems* section can be solved using free body diagrams and Newton’s three laws.

*Equilibrium Conditions*

**Translational equilibrium** exists only when the vector sum of all of the forces acting on an object is zero. This is called the **first condition of equilibrium**, and it is merely a reiteration of Newton’s first law. Remember that when the resultant force upon an object is zero, the object will not accelerate; that may mean that the object is stationary, but it could just as well mean that the object is moving with a constant nonzero velocity. Thus, an object experiencing translational equilibrium will have a constant velocity: both a constant speed (which could be a zero or nonzero value) and a constant direction.

**KEY CONCEPT**

If there is no acceleration, then there is no net force on the object. This means that any object with a constant velocity has no net force acting on it. However, just because the net force equals zero does not mean the velocity equals zero.

**Example:**

Two blocks are in static equilibrium, as shown below:

If block A has a mass of 15 kg and the coefficient of static friction between block A and the surface is 0.2, what is the maximum mass of block B?

**Solution:**

Start by making a free body diagram of each block:

Both blocks have a net force of zero because they are in equilibrium. Therefore, the magnitude of **T** is equal to that of **F**_{g,B}. Asking for the maximum mass of block B means that the force of static friction is maximized (*f*_{s} = *μ*_{s}*N*); further, because block A is in equilibrium, **f**_{s} is equal in magnitude to **T** and **F**_{g,A} is equal in magnitude to **N**. Therefore:

*f*_{s} = *T* and *T* = *F*_{g,B}

ROTATIONAL EQUILIBRIUM

**Rotational motion** occurs when forces are applied against an object in such a way as to cause the object to rotate around a fixed pivot point, also known as the **fulcrum**. Application of force at some distance from the fulcrum generates **torque** (** τ**) or the

**moment of force**. The distance between the applied force and the fulcrum is termed the

**lever arm**. It is the torque that generates rotational motion, not the mere application of the force itself. This is because torque depends not only on the magnitude of the force but also on the length of the lever arm and the angle at which the force is applied. The equation for torque is a cross product:

** τ** =

**r**×

**F**=

*rF*sin

*θ*

**Equation 1.24**

where *r* is the length of the lever arm, *F* is the magnitude of the force, and *θ* is the angle between the lever arm and force vectors.

**KEY CONCEPT**

Remember that sin 90° = 1. This means that torque is greatest when the force applied is 90 degrees (perpendicular) to the lever arm. Knowing that sin 0° = 0 tells us that there is no torque when the force applied is parallel to the lever arm.

*Equilibrium Conditions*

**Rotational equilibrium** exists only when the vector sum of all the torques acting on an object is zero. This is called the **second condition of equilibrium**. Torques that generate clockwise rotation are considered negative, while torques that generate counterclockwise rotation are positive. Thus, in rotational equilibrium, it must be that all of the positive torques exactly cancel out all of the negative torques. Similar to the behavior defined by translational equilibrium, there are two possibilities of motion in the case of rotational equilibrium.

Either the object is not rotating at all (that is, it is stationary), or it is rotating with a constant angular velocity. The MCAT almost always takes rotational equilibrium to mean that the object is not rotating at all.

**Example:**

A seesaw with a mass of 5 kg has one block of mass 10 kg two meters to the left of the fulcrum and another block 0.5 m to the right of the fulcrum, as shown below.

If the seesaw is in equilibrium, find the mass of block 2 and the force exerted by the fulcrum.

**Solution:**

If the seesaw is balanced, this implies rotational equilibrium. Therefore, the positive (counterclockwise) torque exerted by block 1 is equal in magnitude to the negative (clockwise) torque exerted by block 2. Use the fulcrum as the pivot point; this way, both the normal force and the weight of the seesaw will be eliminated from the equation because their lever arms are 0.

To find the normal force exerted by the fulcrum, consider that the seesaw is not only in rotational equilibrium but also translational equilibrium. Therefore, the combined weight of the seesaw and blocks (pointing down) is equal in magnitude to the normal force (pointing up):

**MCAT Concept Check 1.7:**

Before you move on, assess your understanding of the material with these questions.

1. Can a moving object be in equilibrium? Why or why not?

2. If you have an object three times as heavy as you can lift, how could a lever be used to lift the object? Where would the fulcrum need to be placed?