Mechanics help please!

Q) Is my solutions correct both numerically and the forces shown on the diagram?
Have I resolved them correctly or?

I would appreciate it if someone could clarify if I made any mistakes resolving. Thanks! :smile:

Would the weight resolved down into the plane be 750sin(theta) and the weight into the plane be 750cos(theta)
which is the opposite if theta was to the horizontal?

R2 would be R2cos(theta) up the plane and R2sin(theta) into the plane.

Fmax down the plane = Fcos(theta) and up the plane Fsin(theta)

(edited 7 years ago)

Reply 1

ghostwalker

Original post by XxKingSniprxX

Can't say that I like the wording of this question you've been given.

a) At this stage you've not been told that the ladder is in limiting eqilibrium. So it won't necessarily be the case that

F=\mu R

, it could be less.

b) But they expect you to work out the value of each force, so you must, and have, assumed it's limiting equilibrium - fine.

c) You've made a slip here, as you've not taken into account the distance of the forces from the pivot.

20\times R_2\times \cos\theta - 10\times 750\times\sin\theta=0

I'm struggling to interpret what you mean here - you've got "weight resolved down into the plane" and "weight into the plane" :confused:

I suggest talking about components parallel to and perpendicular to the ladder.

Reply 2

XxKingSniprxX

Original post by ghostwalker

Can't say that I like the wording of this question you've been given.

a) At this stage you've not been told that the ladder is in limiting eqilibrium. So it won't necessarily be the case that

F=\mu R

20\times R_2\times \cos\theta - 10\times 750\times\sin\theta=0

I'm struggling to interpret what you mean here - you've got "weight resolved down into the plane" and "weight into the plane" :confused:

I suggest talking about components parallel to and perpendicular to the ladder.

I know .. I hate the wording myself as you have to make assumptions etc.
Q) Is my diagram drawn for (a) sufficient for 5 marks or?
Q) Is the answers I've done for (b) fine, where you just more or less resolving vertical and horizontal?

(c) damn, can't believe I forgot the distance :facepalm:

so if I redid that the angle theta = 38.7 (3sf) is correct?

Yeah, I couldn't exactly figure out if I resolved the component of the forces parallel and perpendicular correctly or? :s-smilie:

Thanks a lot for the help! :hugs:

(edited 7 years ago)

Reply 3

Muttley79

This is a strange question where does it come from?

For part (a) I would use F for the friction as you must not assume it is limiting.

In (b) you can't calculate the forces unless you assume it's in equilibrium which is very unusual ...

As noted, you must include a distance when taking moments.

Reply 4

ghostwalker

Original post by XxKingSniprxX

I know .. I hate the wording myself as you have to make assumptions etc.
Q) Is my diagram drawn for (a) sufficient for 5 marks or?

Can't speak for whoever set the question. A couple of minor comments:

a) You've drawn a vertical arrow at the contact with the wall, implying a frictional force there, which, as you're clearly aware, there isn't.

b) I wouldn't draw the ladder at rougthly 45 degrees. It makes the diagram appear to have some symmetry, and has the potential to lead to false assumptions if you're not careful. Just something to be aware of.

Q) Is the answers I've done for (b) fine, where you just more or less resolving vertical and horizontal?

For some reason I can't see part (b) of the question, but if I recall it said write the values on the diagram. Aside from that it's fine.

so if I redid that the angle theta = 38.7 (3sf) is correct?

Yeah, I couldn't exactly figure out if I resolved the component of the forces parallel and perpendicular correctly or? :s-smilie:

Thanks a lot for the help! :hugs:

You have your sines and cosines correct in part c), and I agree with the final answer.