## MCAT Physics and also Math Review

## Chapter 1: Kinematics and Dynamics

### 1.7 mechanical Equilibrium

So much we’ve to be paying fist to kinematics and also the special situations of linear and projectile motion. However, many times the MCAT will call for us to eliminate acceleration, or otherwise keep a system in equilibrium. To accomplish this, we need to be familiar with examining forces, especially with totally free body diagrams, as well as with the special conditions for translational and rotational equilibrium. The examine of forces and torques is called **dynamics**.

You are watching: For a system in mechanical equilibrium the resultant

FREE human body DIAGRAMS

While we all have actually an intuitive sense of forces (and your effects) in day-to-day life, students frequently struggle to represent them diagrammatically. Illustration **free body diagrams** takes part practice yet will be a an useful tool ~ above the MCAT. On check Day, make sure to draw a free body diagram for any problem in i m sorry you must perform calculations ~ above forces.

**MCAT EXPERTISE**

When managing dynamics questions, always draw a quick picture of what is keep going in the problem; this will keep whatever in its proper relative position and aid prevent girlfriend from making an easy mistakes.

**Example:**

Three human being are pulling ~ above ropes tied to a tires with pressures of 100 N, 150 N, and also 200 N as displayed below. Find the magnitude and also direction the the result force. (Note: sin 30° = 0.5, sin 60° = 0.866)

**Solution:**

First, attract a free body diagram that reflects the forces acting on the tire. Its purpose is come identify and also visualize the exhilaration forces.

The resultant force is merely the sum of the forces. To discover the resultant pressure vector, we require the sum of the force components, presented below.

The net component vectors are displayed graphically below.

The magnitude deserve to then be determined using the Pythagorean theorem:

The direction can be figured out using the tangent function:

Note that this calculation of an inverse tangent is beyond the border of check Day math. The result vector is shown below.

TRANSLATIONAL EQUILIBRIUM

Translational movement occurs when forces cause an item to relocate without any type of rotation. The most basic pathways may be linear, such as when a son slides down a snowy hill ~ above a sled, or parabolic, together in the case of a cannonball shot the end of a cannon. Any problem about translational activity in the *Chemical and also Physical structures of organic Systems* section deserve to be solved using complimentary body diagrams and also Newton’s three laws.

*Equilibrium Conditions*

**Translational equilibrium** exist only when the vector sum of every one of the forces acting on an object is zero. This is dubbed the **first problem of equilibrium**, and it is just a reiteration the Newton’s an initial law. Psychic that when the resultant pressure upon things is zero, the object will not accelerate; that may mean that the thing is stationary, however it can just too mean that the object is moving with a constant nonzero velocity. Thus, an item experiencing translational equilibrium will have actually a constant velocity: both a consistent speed (which could be a zero or nonzero value) and also a consistent direction.

**KEY CONCEPT**

If over there is no acceleration, climate there is no net pressure on the object. This way that any type of object through a constant velocity has actually no net pressure acting ~ above it. However, just because the network force amounts to zero walk not typical the velocity equals zero.

**Example:**

Two blocks space in static equilibrium, as presented below:

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

**Solution:**

Start by do a complimentary body chart of every block:

Both blocks have a net pressure of zero due to the fact that they room in equilibrium. Therefore, the magnitude of **T** is equal to that of **F**g,B. Questioning for the maximum massive of block B means that the force of revolution friction is maximized (*f*s = *μ*s*N*); further, due to the fact that block A is in equilibrium, **f**s is same in size to **T** and **F**g,A is same in magnitude to **N**. Therefore:

*f*s = *T* and also *T* = *F*g,B

ROTATIONAL EQUILIBRIUM

**Rotational motion** wake up when pressures are applied versus an object in such a method as to reason the thing to rotate roughly a resolved pivot point, additionally 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 the generates rotational motion, not the mere applications of the force itself. This is because torque counts not only on the size of the force but likewise on the size of the bar arm and the angle at i beg your pardon the pressure is applied. The equation because that torque is a cross product:

** τ** =

**r**×

**F**=

*rF*sin

*θ*

**Equation 1.24**

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

**KEY CONCEPT**

Remember the sin 90° = 1. This method that speak is best when the force applied is 90 degrees (perpendicular) come the bar arm. Discovering that sin 0° = 0 tells united state that over there is no torque as soon as the force used is parallel come the bar arm.

*Equilibrium Conditions*

**Rotational equilibrium** exists only when the vector amount of every the torques exhilaration on an item is zero. This is dubbed the **second problem of equilibrium**. Torques that generate clockwise rotation are thought about negative, while torques that generate counterclockwise rotation room positive. Thus, in rotational equilibrium, it should be that every one of the hopeful torques precisely cancel out every one of the an adverse torques. Similar to the behavior defined through translational equilibrium, there space two possibilities of motion in the instance of rotational equilibrium.

Either the object is no rotating at all (that is, that is stationary), or it is rotating with a constant angular velocity. The MCAT almost always takes rotational equilibrium to average that the thing is not rotating at all.

**Example:**

A seesaw v a massive of 5 kg has actually one block of massive 10 kg 2 meters come the left of the fulcrum and also another block 0.5 m to the ideal of the fulcrum, as shown below.

If the seesaw is in equilibrium, discover the mass of block 2 and also the pressure exerted by the fulcrum.

**Solution:**

If the seesaw is balanced, this indicates rotational equilibrium. Therefore, the optimistic (counterclockwise) talk exerted by block 1 is equal in magnitude to the an adverse (clockwise) torque exerted through block 2. Usage the fulcrum as the pivot point; this way, both the common force and the load of the seesaw will be eliminated from the equation because their lever arms are 0.

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To discover the normal force exerted by the fulcrum, consider that the seesaw is not just in rotational equilibrium but additionally translational equilibrium. Therefore, the merged weight that the seesaw and blocks (pointing down) is equal in magnitude to the normal pressure (pointing up):

**MCAT Concept inspect 1.7:**

Before you relocate on, evaluate your knowledge of the material with this questions.

1. Deserve to a relocating object be in equilibrium? Why or why not?

2. If you have things three time as hefty as you have the right to lift, how could a lever be used to background the object? where would the fulcrum need to be placed?