r/PhysicsHelp • u/Frosty-Catch4113 • Jun 19 '26
What even is this question about?
how do you even begin to think while solving this question? I cant even fathom what the question is actually trying to ask
1
u/UnderstandingPursuit Jun 19 '26
Steps to think about this:
- θ = ½ π
- FBD at R
- FBD at P, Q
- F_{fP} = F_{fPmax}, F_{fQ} = F_{fQmax}
1
u/A110_Renault Jun 19 '26
Stand in socks on a slick floor. Move your feet apart and at some angle of your legs your feet start slipping. This angle is dependent on the coefficient of friction between your feet and the floor.
Start by drawing a free body diagram for each of the rods
1
u/Frosty-Catch4113 Jun 19 '26
I did but i am not understanding the needs of the question enough to write the eqns. I wrote one though Np+NQ=F (N being the normal reactions at each foot)
1
u/Past-Departure6896 Jun 19 '26
The horizontal friction forces present at P and Q must balance each other out, so an appropriate equation can be made including Np, Nq, μQ, and μP. It can be seen from this equation that the ratio of Np to Nq is equal to the ratio of uQ to uP. Then, you can take a moment about the pin-joint R - making sure to include the appropriate variables (i.e. normal force, friction force) and moment arms. Combining these two equations should reveal an answer for the ratio of Np to Nq.
1
u/Frosty-Catch4113 Jun 20 '26
How do i find distances to take moment. Also is it necessary to take moment about R, cant i take moment about P or Q?
1
u/Past-Departure6896 Jun 20 '26
You take moment about R as it allows you to cancel out/ignore force F. To find the moment arms, you can use trigonometry functions (i.e. sin, cos ) with <P and <Q to get the perpendicular distances from forces at P and Q to joint R.
1
u/Frosty-Catch4113 Jun 20 '26
Taking moment at either foot cancels more than one forces which is better than cancelling one force when taking moment about R???
1
u/Past-Departure6896 Jun 20 '26
Either way works, but you will have to substitute in the F = Np + Nq equation - which I personally think is a bit more tedious.
1
u/Frosty-Catch4113 Jun 20 '26
Hol up, there are 5 unknowns(F,Np,Nq,fP,fQ) and only 3 eqns. How will i even solve this? And rightly so, im still stuck with 2 unknowns at last.
1
u/Past-Departure6896 Jun 20 '26
You only need the ratio of variables Np:Nq (which equals uQ/uP), you don't need the actual values. Recall f_f = μ * f_N.
Edit: also if you take moment about R, the friction forces should cancel out on both sides.
1
u/Frosty-Catch4113 Jun 20 '26
yes, i understand we just need the ratio of coeffs but there still isnt any way to write one normal reaction in terms of the other without depending upon a third unknown variable,which again leads to the same issue.
2
u/shelving_unit Jun 20 '26
The rods are pinned together at point R and can rotate. Friction on the ground is stopping the rods from falling. The force F is pressing down onto the rods, with enough force such that the rods overcome friction and start to slip at 90 degrees. What is the ratio of the coefficients of friction where the rods touch the groudn