The Student Room Group

Connected Particles, different coef of friction

Hi again,
Second question I'd like to put up here for perusal by the collective brains of the outfit. As mentioned, they all originate from the Edexcel Yr2 Stats and Mechs Practice book. Sadly, the book doesn't provide full solutions just the bare answer.

Please refer to attached.

Part a. is straight forward a = 10/9 m/s/s.

Part b. Has got me stuck.

Really appreciate it if any of you folks can keep me honest and set me straight.
Original post by veem461
Hi again,
Second question I'd like to put up here for perusal by the collective brains of the outfit. As mentioned, they all originate from the Edexcel Yr2 Stats and Mechs Practice book. Sadly, the book doesn't provide full solutions just the bare answer.

Please refer to attached.

Part a. is straight forward a = 10/9 m/s/s.

Part b. Has got me stuck.

Really appreciate it if any of you folks can keep me honest and set me straight.


What have you tried? Where are you stuck?
Reply 2
Hi there,
Many thanks for writing.

Referring to attached (initial attempt) - as I mentioned, I've become bogged down setting things up for part b. I suspect I've already come off the rails in what I have done.

So, given the connected particles are in motion they therefore have overcome Limiting Equilibrium and hence Frictional Forces (Fr3 and Fr4) both = mu * their Reaction Forces (R1 and R6) resp.

Also, given coef friction for P is mu and for Q is 2mu then Fr3 = 3g(mu) and Fr4 = 8g(mu)

However, I am concerned that I have not taken into account the vertical component of F8 = 45sin(40).

At which, I am stuck. If only these Edexcel boffins would have provided full solutions (as they do in the course collateral) I would have had a fighting chance on figuring this out. Ive done innumerable questions on connected particles but never one like this i.e. different coef of friction for connected particles. These sort of questions have tended to be truck and trailer or train and carriage type scenarios.

Apologies for not being able to have dug deeper. I hope this helps explain how bogged I am.

VBR
Original post by veem461
Hi there,
Many thanks for writing.

Referring to attached (initial attempt) - as I mentioned, I've become bogged down setting things up for part b. I suspect I've already come off the rails in what I have done.

So, given the connected particles are in motion they therefore have overcome Limiting Equilibrium and hence Frictional Forces (Fr3 and Fr4) both = mu * their Reaction Forces (R1 and R6) resp.

Also, given coef friction for P is mu and for Q is 2mu then Fr3 = 3g(mu) and Fr4 = 8g(mu)

However, I am concerned that I have not taken into account the vertical component of F8 = 45sin(40).

At which, I am stuck. If only these Edexcel boffins would have provided full solutions (as they do in the course collateral) I would have had a fighting chance on figuring this out. Ive done innumerable questions on connected particles but never one like this i.e. different coef of friction for connected particles. These sort of questions have tended to be truck and trailer or train and carriage type scenarios.

Apologies for not being able to have dug deeper. I hope this helps explain how bogged I am.

VBR

I've not gone through your attachment, but based on what you've written above:

1) You are right to suspect that the reaction force for Q is affected by the vertical component of the 45N force.

2) To proceed, you should note that the net vertical force on Q must be zero (as it is not moving at all in the vertical direction).

3) The vertical forces acting on Q are 4g (downwards), R6 (upwards) and 45sin(40) upwards. These must sum to zero, which gives you a way to find R6.
(edited 1 year ago)
Reply 4
Many thanks. Will give this another crack.

Quick Reply

Latest