The Student Room Group
Competition - post a photo you've taken of nature >>
Competition - post a photo you've taken of nature >>
Competition - post a photo you've taken of nature >>

Better explanation of mechanics question

Hi all,
I've got this question here, but I'm really very confused as to what it's actually asking me to do. Could anyone explain it a little better for me?
IMG_0253.JPG
Thanks in advance
Reply 1
Avatar for thomoski2
thomoski2
OP
5
Just encountered a second question with the same issue. What is v here? Is it a position vector?
IMG_0254.JPG
Original post by thomoski2
Hi all,
I've got this question here, but I'm really very confused as to what it's actually asking me to do. Could anyone explain it a little better for me?
IMG_0253.JPG
Thanks in advance


This one just wants you to use F=ma\mathbf{F}=m \mathbf{a} to show that a=(x¨y¨z¨)=(00−g)\mathbf{a}= \begin{pmatrix} \ddot{x} \\ \ddot{y} \\ \ddot{z} \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \\ -g \end{pmatrix}. It should be a one liner type of answer.
Original post by thomoski2
Just encountered a second question with the same issue. What is v here? Is it a position vector?


I think that the equation is supposed to read v=x˙i+z˙k\mathbf{v} = \dot{x} \mathbf{i} + \dot{z} \mathbf{k} which is obviously the velocity vector and the question has a typo. Not sure other than that, it's not very clear.

In which case, all you need to do is solve x¨=λ\ddot{x} = \lambda and z¨=μz˙\ddot{z} = \mu \dot{z} and hence determine x˙,z˙\dot{x}, \dot{z}.
Reply 4
Avatar for thomoski2
thomoski2
OP
5
Original post by RDKGames
This one just wants you to use F=ma\mathbf{F}=m \mathbf{a} to show that a=(x¨y¨z¨)=(00−g)\mathbf{a}= \begin{pmatrix} \ddot{x} \\ \ddot{y} \\ \ddot{z} \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \\ -g \end{pmatrix}. It should be a one liner type of answer.

How do I go about doing that?
Reply 5
Avatar for thomoski2
thomoski2
OP
5
Original post by thomoski2
How do I go about doing that?


RDKGames?
Original post by thomoski2
How do I go about doing that?


Sorry, forgot to reply to this.

Well this is the very basic question that your theory should answer.

If you drop a stone from a height, then under a gravitational field it will have a gravitational force that drags it down. What is this force as a vector?

Likewise, since there are not other forces acting on it, this must be equal to the mass of the particle multiplied by its acceleration.
Reply 7
Avatar for thomoski2
thomoski2
OP
5
Original post by RDKGames
Sorry, forgot to reply to this.

Well this is the very basic question that your theory should answer.

If you drop a stone from a height, then under a gravitational field it will have a gravitational force that drags it down. What is this force as a vector?

Likewise, since there are not other forces acting on it, this must be equal to the mass of the particle multiplied by its acceleration.


So how do I apply this to answer the question? Sorry for all this, I had no lectures on this topic due to the strikes. I get that -mg=mz: but I don’t understand the part about position vector
(edited 6 years ago)
Original post by thomoski2
So how do I apply this to answer the question? Sorry for all this, I had no lectures on this topic due to the strikes. I get that -mg=mz: but I don’t understand the part about position vector


You don't need to do anything with it. The question is just asking you to show that the height at time tt, which is defined by z(t)z(t), satisfies the ODE z¨=−g\ddot{z} = -g. It does not ask you to do anything with z(t)z(t) explicitly.

Your eq. is wrong, you should get −mg=mz¨-mg = m \ddot{z} instead - note that z¨\ddot{z} is the second derivative of zz with respect to time. Then answer is one step away from here on in, hence why it's a one liner.
(edited 6 years ago)

Quick Reply

Related discussions

Latest

Trending

Trending

Articles for you