How To Find Mass With Acceleration And Force

The Mighty F = ma

This equation is one of the most useful in classical physics. Information technology is a concise statement of Isaac Newton'due south Second Law of Movement, property both the proportions and vectors of the Second Law. It translates every bit:

The net forcefulness on an object is equal to the mass of the object multiplied by the acceleration of the object.

Or some simply say:

Force equals mass times acceleration.

We will talk over the following about this formula:

• Units for force, mass, and acceleration
• Calculations and algebra
• Unit of measurement relationships
• Self-test problems
• Vector directions
• Directly and inverse proportions
• Relationship to kinematics formulas

Units for strength, mass, and dispatch

Force is measured in Newtons, N. Mass is measured in kilograms, kg. Acceleration is measured in meters per 2d squared, m/s2.

When you multiply a kilogram (mass unit) times a meter per second squared (acceleration unit) you lot get a kilogram-meter per second squared.

And so a unit for forcefulness is actually the kilogram-meter per 2nd squared. Notwithstanding, no i really says that. The unit for force is named subsequently Isaac Newton, and it is called the 'Newton', abbreviated 'N'. Ane Newton is one kilogram-meter per second squared.

Another almost identical way to call up almost the force unit of measurement is that one Newton is the size of a strength needed to accelerate a mass of one kilogram at a rate of one meter per second squared, every bit in:

Calculations and algebra

Here is a slideshow that shows three bones issues using F=ga. Also presented is the algebra used to rearrange the formula when solving for mass or acceleration.

Unit relationships

Later going through the higher up slideshow, one might wonder how the units for these equations resolve. For example, the following is true:

a = F / m

Only how does an dispatch unit of measurement (on the left) go the same as a force unit divided by a mass unit of measurement (on the correct)? All of this is explained in the post-obit scrolling panels.

Self-examination problems

Here are some problems that work with basic F=ma algebra. All of that is explained in the 'Calculating with F=ma' sit-in above. Each trouble page has a back button that volition return y'all to this page.

Vector directions

Here we will show that in the equation F=granda the acceleration vector, a, has the same direction as the net force vector, F.

First, recall that when nosotros multiply a scalar times a vector, the effect is a vector that has the same direction equally the original. And then if nosotros multiply scalar 2 times vector P we get a vector as the result which has the same direction every bit vector P. We could requite this result vector a name, say Q. This is all shown in the post-obit animation:

Now let's see how all this works out with the F and the a vector in the equation F=thoua. Note that the right side of the equation is mass times acceleration. Mass is a scalar, and acceleration is a vector. So the right side of this equation is a scalar times a vector. This multiplication yields a vector that is called the forcefulness vector, or F, which is on the left side of the equation.

As above with P and Q, vector a is in the aforementioned direction as vector F. Therefore, the dispatch of an object is in the same direction of the applied net forcefulness. Here is another animation showing all of this for the a and F vectors:

And and then we have shown that the formula F=ma contains the information that an object'southward dispatch vector is aimed in the same management as its practical cyberspace force vector.

Direct and changed proportions

In that location is a direct proportion between acceleration and practical cyberspace force.

Data about direct proportions can be found here.

Let'southward see if the equation F=ma contains the direct proportion between force, F, and dispatch, a. That is, does F=ma contain the data that acceleration is directly proportional to the applied net strength?

If acceleration is directly proportional to the applied net force, then by whatever factor acceleration changes, force changes by the same factor. To see this nosotros will need to consider an object with constant mass. Let'south consider an object with a mass of 5 kg.

Suppose this object has an acceleration of 3 g/sⁱⁱ. Let's call this a₁. And so:

a1 = three m/s2

What strength is necessary to cause this acceleration to our 5 kg object? Let'southward effigy that out and call it F₁:

F1 = maane Fi = (5 kg)(3 m/s2) Fane = 15 North

Now allow's think about this object when it is moving at twice this acceleration. That is, we will change the acceleration past a factor of two. Allow's phone call this new acceleration a_two. So:

a2 = 2(aone) = 2(3 m/s2) = 6 thousand/s2

What force is now necessary to crusade this new dispatch, a₂? Nosotros will call this force F_two, and here is its calculation:

F2 = ma2 F2 = (v kg)(6 m/s2) F2 = 30 N

Now that second force, thirty N, is twice the get-go strength, 15 N:

F₂ = 2F_i

xxx N = 2(15 N)

xxx N = 30 N

And that second dispatch, half-dozen thou/due south², is twice the first acceleration, 3 m/s²:

a_ii = 2(a_i)

half-dozen 1000/s^two = two(iii one thousand/south²)

half dozen m/south^two = 6 m/s²

Conspicuously, both the acceleration and the force on this object change by the same cistron, two. These matching gene changes would occur for any cistron alter and for any constant mass. So the equation F=ma contains the direct proportion between dispatch and the practical net strength.

The inverse proportion betwixt acceleration and mass

Now, what almost the inverse proportion between acceleration and mass, is that contained inside F=ma?

Information about inverse proportions can be found here.

Allow's consider two situations, each with the same applied internet force, say 6 Due north. That is, object 1 volition have a internet strength of six N applied to information technology, then will object 2.

Nosotros will say our offset object has a mass of three kg. Then:

grandi = 3 kg

What volition be the acceleration of this object? Allow'south call that dispatch a₁, and let'southward solve for it using F=ma. Then:

F = thousand_onea₁

Rearranging and calculating the in a higher place equation for dispatch we get:

aane = F / one thousand1 ai = half-dozen Due north / 3 kg a1 = ii one thousand/southward2

At present let'due south cutting the mass to 1-tertiary. That is, we will consider a alter in mass past a cistron of ane/3. That would brand the mass of object two to be 1 kg, or grand_two = i kg. If we apply the aforementioned force, six North, to object 2, what will exist its acceleration? Hither'south the calculation:

a2 = F / m2 a2 = 6 N / one kg a2 = 6 thou/southwardtwo

Then the second mass is 1/3 the first mass, since:

1000_ii = (i/3)m₁

1 kg = (one/3)(3 kg)

1 kg = 1 kg

And the second acceleration is three times the kickoff acceleration, as in:

a₂ = (3)a_ane

six yard/southⁱⁱ = (3)(2 1000/s²)

6 m/s² = vi m/s^two

Conspicuously, the dispatch and mass change by reciprocal (or inverse) factors. The factors are 3 and 1/iii, respectively. Since these two quantities change past reciprocal (inverse) factors, these ii quantities are in an inverse proportion.

There is really nothing special almost our choice of factor changes here. Try changing the mass by a factor of 5 and calculate to come across the dispatch change past a gene of 1/5.

Looks like the formula F=ma contains both the direct and inverse proportions we described in our primary Newton's Second Police of Motion folio.

Relationship to kinematics formulas

Let's come across how F=ma works itself into the kinematics formulas for accelerated movement, effectively changing kinematics into dynamics.

Data about equations for accelerated motion can be establish here.

These equations are presented here with the subscript 'i' meaning 'initial'. So this symbol:

v _i

Would mean 'velocity initial' or 'the initial velocity'.

Now you are just every bit probable to come across the subscript 'o' meaning 'original' used for exactly the aforementioned concept, the first, initial, or original. This symbol:

v_o

Would mean 'velocity original' or 'the original velocity'.

So the subscripts 'i' and 'o' carry the aforementioned concept:

v_i is the aforementioned every bit 5_o

Hopefully the above volition clear up any problems over the mode that these equations are presented on the Equations for Accelerated Motion page and the style they are presented here. Example problems follow.

As it turns out, since:

a = F / yard

Knowing an object's mass and the internet force on it is every bit good as knowing the dispatch of the object. Divide the net force by the mass, and you volition find the acceleration of the object. In one case the acceleration is known, you can utilise the kinematic equations to discover the motion which the object will undertake. With all of this you tin figure out where things are going and when they will get there. Cheers, Isaac Newton.

Permit'due south note the substitutions in the following kinematics equations.

d = five_it + (one/2)atⁱⁱ

Displacement equals the initial (or original) velocity multiplied by time plus one-half times the acceleration multiplied past fourth dimension squared.

5_f = v_i + at

The last velocity is equal to the initial velocity plus the acceleration multiplied by time.

v_f ² = v_i ² + 2ad

The square of the final velocity is equal to the square of the initial velocity plus ii times the acceleration times the displacement.

Instance dynamics issues

Post-obit are some example problems showing how the above exchange allows us to understand the motion of an object and the forces causing its motion. The parameters for each example can exist modified with the 'less' and 'greater' buttons. Adjust them to meet how the calculations work in several situations.

And so, to sum information technology up for F=ma:

• It'southward a concise and sufficient argument of Newton's Second Police force of Movement.
• In vector form (F=ma, bold quantities are vectors) it shows that acceleration and the applied internet force are parallel vectors.
• Keeping the mass constant demonstrates a direct proportion between acceleration and the applied net force.
• Keeping the force constant demonstrates an inverse proportion betwixt dispatch and mass.

And of form, y'all can navigate with the kinematics equations. Only have your mass and internet force handy so that you can calculate the dispatch.

Source: http://zonalandeducation.com/mstm/physics/mechanics/forces/newton/mightyFEqMA/mightyFEqMA.html

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