M1 jan 2008

Hey.
For Part 6 (iii). Im not sure how i meant to do the question such that i can "justify my answer fully". Whats the reason for this wording?
thanks

Reply 1

username1214456

Posted from TSR Mobile

I'm guessing you have to show that the horizontal component of the force is bigger than Fmax
And so justify how that makes it move

(edited 8 years ago)

Reply 2

Notnek

Original post by SamuelN98

Hey.
For Part 6 (iii). Im not sure how i meant to do the question such that i can "justify my answer fully". Whats the reason for this wording?
thanks

If the horizontal component is less than

\mu R

then friction will be enough to oppose motion i.e. the friction will be equal to the force.

If the force is

\geq \mu R

then friction will have reached it's limit and so the frictional force will be equal to

\mu R

(edited 8 years ago)

Reply 3

username1214456

Posted from TSR Mobile

Mate I know how to do it lol
The OP doesn't

Reply 4

Zacken

Original post by Sharaf371

Mate I know how to do it lol
The OP doesn't

Are you sure?

Original post by Sharaf371

I'm guessing you have to show that the horizontal component of the force is bigger than Fmax

It's not bigger... it's smaller.

And so justify how that makes it move

It doesn't move, that's the whole point.

We tend to be taught

F = \mu R

in mechanics - but the actual expression or formula is

F \leq \mu R

Reply 5

Notnek

Original post by Sharaf371

Posted from TSR Mobile

Mate I know how to do it lol
The OP doesn't

Sorry - quoted the wrong person!

Reply 6

username1214456

Posted from TSR Mobile

Original post by Zacken

Are you sure?

It's not bigger... it's smaller.

It doesn't move, that's the whole point.

We tend to be taught

F = \mu R

in mechanics - but the actual expression or formula is

F \leq \mu R

I haven't done the question so I was "guessing". Obviously if I did do it then I'd find if it was bigger or smaller. And so I would know it moves or not