63 Accelerating the Body

Newton’s Second Law of Motion

tells us that we need a in order to create an .   As you might expect, a larger net force will cause a larger acceleration, but the more matter you are trying to accelerate the larger force will be required. summarizes all of that into a single equation relating the , , and :

(1)   \begin{equation*} \bold{F_{net}} = m\bold{a} \end{equation*}

Finding Net Force from Acceleration

Everyday Example: Parachute Opening

In the previous chapter we found that if opening a parachute slows a skydiver from 54 m/s to 2.7 m/s in just 2 s of time then they experienced an average upward acceleration of 26 m/s/s . If the mass of our example skydiver is 85 kg, what is the  average net force on the person? What is the average force on them from the harness?

We start with Newton’s Second Law of Motion

(2)   \begin{equation*} \bold{F_{net}} = m\bold{a} \end{equation*}

Enter in our values:

(3)   \begin{equation*} \bold{F_{net}} = (85\,\bold{kg})(26\,\bold{m/s/s}) = 2200 <strong>N</strong> \end{equation*}

The person experiences an average net force of 2200 N upward during chute opening. When the chute begins to open they are still moving near terminal velocity so air resistance is nearly balancing their weight and the harness provides most of the extra 2200 N upward force on the person. That force is 2.7 times their body weight (Fg = 85 kg x 9.8 m/s/s = 833 N).

Reinforcement Exercises: Failed Chute Opening

If the skydiver in the previous example experienced a failed chute opening and hit the ground so that they came to an abrupt stop in a time of  only 0.2 s (instead of slowing to 2.7 m/s in 2 s) what is their acceleration?

What is the average net force on them during the stop?

The normal force from the ground must be large enough to cancel the skydiver’s weight and still provide the upward net force you found above. How big is the normal force from the ground?

How many times larger is this normal force than their weight?

Reinforcement Exercises: Baby Toss

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You’d like to accelerate a 7.6 kg baby from rest to 1.5 m/s  over 1.0 s. What is the baby’s acceleration?

What net force is required?

Draw a free body diagram of the situation, that shows the baby’s and the force you are providing.

Considering the baby’s weight, what actual force do you need to provide for the baby to experience the net force you calculated above?

Once the baby is in the air, what acceleration will they have? (Assume they are moving slowly enough that air resistance is ).

Using the definition of average acceleration, find the time that the baby was in the air before their velocity reached zero at the top of the toss.

What was the total “hang time” for the baby? (Finish this unit to learn how we calculate the height reached by the baby in this case).

When you catch the baby do you keep your arms held rigid or do you move your hands downward with the baby as you make the catch? Explain why in terms of .

Finding Acceleration from Net Force

If we know the net force and want to find the acceleration, we can solve in terms of the acceleration instead:

(4)   \begin{equation*} \bold{a} = \frac{\bold{F_{net}}}{m} \end{equation*}

Now we see that larger create larger and larger masses reduce the size of the acceleration. In fact, an object’s is a direct measure of an objects resistance to changing its motion, or its .

Reinforcement Exercises

You slide a box across the floor by applying a 220 N force to the right. applies a reactive 170 N force on the box to the left.

What is the size and direction of the on the box?

The box has a mass of 25 kg. What is the size and direction of the on the box?

The box is not accelerating in the vertical direction, so what is the net vertical force? [Hint: Forces in different dimensions (vertical and horizontal) don’t affect one another and applies separately to each dimension.]

How big is the on the box?

What is the value of the kinetic?

Check out this simulation to see how forces combine to create net forces and accelerations:

Forces and Motion: Basics

Free-Fall Acceleration

In the absence of air resistance, heavy objects do not fall faster than lighter ones and all objects will fall with the same acceleration. Need experimental evidence? Check out this video:

It’s an interesting quirk of our universe that the same property of an object, specifically its , determines both the on it and its resistance to accelerations, or . Said another way,  the inertial mass and the gravitational mass are equivalent. That is why we the free-fall acceleration for all objects has a of 9.8 m/s/s, as we will show in the following example.

Everyday Example: Free-Falling

Let’s calculate the initial of our example skydiver the moment they jump. At this moment they have the pulling them down, but they have not yet gained any , so the () is zero. The is then just gravity, because it is the only force, so they are in free-fall for this moment. Starting with :

(5)   \begin{equation*} \bold{a} = \frac{\bold{F_{net}}}{m} \end{equation*}

And inserting our known formula for calculating force of gravity near the surface of Earth and including a negative sign because down is our negative direction, (\bold{F_g} = -mg):

(6)   \begin{equation*} \bold{a} = \frac{-mg}{m} \end{equation*}

We see that the cancels out,

(7)   \begin{equation*} \bold{a} = \frac{-\cancel{m}g}{\cancel{m}} = -g = -9.8 \bold{\frac{m}{s^2}} \end{equation*}

We see that our is negative, which makes sense because the acceleration is downward. We also see that the size, or magnitude, of the acceleration is g = 9.8 m/s2. We have just shown that in the absence of , all objects falling near the surface of Earth will experience an  equal in size to  9.8 m/s2, regardless of their and . Whether the free-fall acceleration is -9.8 m/s/s  or +9.8 m/s/s depends on if you chose downward to be the negative or positive direction.

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