Question

Find the value of ax, the x-component of the acceleration and of ay for both of the diagrams. The forces are acting on a 3.0 kg object.

Answer 1

Hello! Do you know the basic technique for these?

Answer 2

not really, i don't know how to start

Answer 3

Okay, I'd like to help!
First, are you comfortable with vector component notation (\(\hat i,\ \hat j\))?
I really don't expect you to be, but it would be convenient, so I'll ask.

Answer 4

yea i think

Answer 5

Eh, if you're not fond of it, I don't have to use it!

Answer 6

So, here's the thing about those forces: each force contributes to the net force, right? Well, that determines the accelaration. Sure, that's the basic \(F=ma\). But! You can also use the \(x\)-component of force to find the \(x\)-component of acceleration with \(F=ma\)!

So, first we need to find the \(x\)- and \(y\)-components of the net force by finding the net \(x\) and \(y\) net forces.


Can you find the net \(x\) force in the first one?

Answer 7

It's \(-2.0{\rm\ N}\) and \(4.0\rm\ N\) put together to get...

Answer 8

@mony01 ?

Answer 9

2N

Answer 10

Yep! So, that is the \(x\)-component of net force. So, using \(F=ma\), what is the \(x\)-component of acceleration?

Answer 11

\(F=2\rm\ N\), \(m=3.0\rm\ kg\)

Answer 12

a=-1

Answer 13

Well, if you divide both sides by \(m\), the \(F=ma\) turns into \(\dfrac Fm=a\).

Want to give that another shot?

Answer 14

2/3

Answer 15

Right!

Now, we got that by first finding the net \(x\) force, and then dividing it by the mass to get the \(x\) acceleration.

Now, why don't you try that with the \(y\) component?

Answer 16

ok I have a question on the first part why was it -2 and 4, and not -4 and 2?

Answer 17

That is honestly a very good and underrated question.

In physics, we care about sign. We make it up, of course, but then it matters. It the difference between up and down, so it is pretty important. What you need to look for is how \(x\) is represented on the axes.

Since \(x\) is on the right side, I assume that \(x\) is positive on that side. So, for \(y\), you can assume the up side is positive.

Answer 18

I double checked with the picture you posted :)

Answer 19

so it is -4 and 2?

Answer 20

Oh, yeah! You're right! Good catch. I was looking at the next one. I had my tabs in the wrong order, so.. woops, hehe!

Answer 21

so a=-2/3 and for the y-component is it 3N?

Answer 22

@theEric

Answer 23

Sorry I was gone!

For the \(x\)-component, you're right! For the \(y\), there is \(\rm 3N\) up and \(\rm3N\) down, so there is NO net force, and so no acceleration in the \(y\) directions.

Answer 24

so its 0

Answer 25

Right! So now you have the \(x\) and \(y\) components of the acceleration, like it asked for! One more force diagram to go!

Answer 26

would the units be in N

Answer 27

For acceleration, you want to the "change in velocity per time." Since velocity is \(\rm m/s\), and time is \(\rm s\), you definitely want to go with \(\rm m/s^2\) as the units for acceleration.

Answer 28

If you think about it, \(F=ma\) and \(m\) has its own units. So the force and the acceleration can't have the same units, or the equation would have a problem.

Answer 29

ok

Answer 30

for the other diagram is the ax=2/3 and ay=0

Answer 31

Yup! :)

Answer 32

Bravo! Make sure you keep the units and I agree completely.

Answer 33

ok thank you very much you were really helpful

Answer 34

I'm glad I could help! You did practically the whole thing by yourself, then! Take care! :)