pulley question (2 Viewers)

[kkk579]

for this q why did they let fnet of m2 be all that? isnt that fnet of the entire system? im so confused... ive also attached my working out below, if anyone could tell me where i went wrong in my w/o that would be great. thank you!!

 

[kkk579]

I didnt add the friction force into tbe diagram but essentially i thought it was in the direction pointing down the slope

 

[kkk579]

someone pls help

 

[Drongoski]

Remember: friction is not a fixed force. I call it an "annoying force" in that if an object is trying to move up the incline, friction acts down the incline to oppose your motion. If the object is sliding down the plane, then the friction will act in the opposite direction, up the incline. And remember the friction is a variable force; can keep increasing, as required, but in this case up to Normal Reaction(of M1 & M2) x 0.251(the maximum friction possible).

But friction, although annoying sometimes, is something we cannot do without. Without friction, we won't be able to walk or drive our car.


Note: correction above.

 

[kkk579]

Remember: friction is not a fixed force. I call it an "annoying force" in that if an object is trying to move up the incline, friction acts down the incline to oppose your motion. If the object is sliding down the plane, then the friction will act in the opposite direction, up the incline. And remember the friction is a variable force; can keep increasing, as required, but in this case up to mass x 0.251(the maximum friction possible).

But friction, although annoying sometimes, is something we cannot do without. Without friction, we won't be able to walk or drive our car.

I thought that the objects would move up the slope because of pulley no?

 

[liamkk112]

I thought that the objects would move up the slope because of pulley no?

depends on the masses, angle of inclination, etc, so the setup of the pulley could cause the objects to move up the slope, or it could cause it them to move down the slope. it this case is seems like the objects will move up the slope, but the question asks for the magnitude of the force, so sign/direction is omitted here

 

[kkk579]

depends on the masses, angle of inclination, etc, so the setup of the pulley could cause the objects to move up the slope, or it could cause it them to move down the slope. it this case is seems like the objects will move up the slope, but the question asks for the magnitude of the force, so sign/direction is omitted here

oh ok yh that makes sense, but why is my w/o wrong?

 

[liamkk112]

oh ok yh that makes sense, but why is my w/o wrong?

for one thing, you said that the tension in the two ropes is the same, which isn't always true, so there should be 2 different tension forces. tension forces also always point away from the mass, so the tension force connecting m1 to m2 is in the wrong direction. also, the friction should be acting down the slope, which is in the same direction as the force pushing the masses down the slope, so no negative sign for the friction if u use the sign convention ur using. i would prob suggest to just redraw the free body diagram lol, hopefully that will fix ur working out, someone else who knows physics better can prob give a better response

 

[kkk579]

for one thing, you said that the tension in the two ropes is the same, which isn't always true, so there should be 2 different tension forces. tension forces also always point away from the mass, so the tension force connecting m1 to m2 is in the wrong direction. also, the friction should be acting down the slope, which is in the same direction as the force pushing the masses down the slope, so no negative sign for the friction if u use the sign convention ur using. i would prob suggest to just redraw the free body diagram lol, hopefully that will fix ur working out, someone else who knows physics better can prob give a better response

The tension in the force connecting m1 and m2 wpuld be pointing away from the point of attachment right? So wouldnt the arrow from m1 side and arrow from m2 side both be pointing towards the middle of the rope?

 

[kkk579]

whyv

for one thing, you said that the tension in the two ropes is the same, which isn't always true, so there should be 2 different tension forces. tension forces also always point away from the mass, so the tension force connecting m1 to m2 is in the wrong direction. also, the friction should be acting down the slope, which is in the same direction as the force pushing the masses down the slope, so no negative sign for the friction if u use the sign convention ur using. i would prob suggest to just redraw the free body diagram lol, hopefully that will fix ur working out, someone else who knows physics better can prob give a better response

why would there be no neg sign for friction? i assigned down the slope as negative abd up thr slope as positive, hence why the weight component and friction is neg and the tension is pos

also ive only learnt that tension is equal everywhere in a rope

 

[liamkk112]

The tension in the force connecting m1 and m2 wpuld be pointing away from the point of attachment right? So wouldnt the arrow from m1 side and arrow from m2 side both be pointing towards the middle of the rope?

yeah but i think u included both the directions in ur calc for m2 (unless if im reading it wrong). in which case they cancel out and u dont account for tension force at all. so when calculating the tension force between m2 and m1 "from m2's perspective" you need to only have the tension force pointing down the slope here

whyv

why would there be no neg sign for friction? i assigned down the slope as negative abd up thr slope as positive, hence why the weight component and friction is neg and the tension is pos

also ive only learnt that tension is equal everywhere in a rope

if the tension between m2 and m1 is down the slope, it has the same sign as friction. then the only positive tension is the one between m2 and m3

 

[kkk579]

yeah but i think u included both the directions in ur calc for m2 (unless if im reading it wrong). in which case they cancel out and u dont account for tension force at all. so when calculating the tension force between m2 and m1 "from m2's perspective" you need to only have the tension force pointing down the slope here

if the tension between m2 and m1 is down the slope, it has the same sign as friction. then the only positive tension is the one between m2 and m3

im kinda confused on what ur tryna say, what i was trying to do first was finding the net force of the entire system and letting that equal (mass*m1+m2+m*)a to find the acceleration. and then i multiplied that acceleration by m2s mass

 

[liamkk112]

im kinda confused on what ur tryna say, what i was trying to do first was finding the net force of the entire system and letting that equal (mass*m1+m2+m*)a to find the acceleration. and then i multiplied that acceleration by m2s mass

ohhh okay i see. idk how u would find the net force on the whole system tho, i was more approaching it from focusing on a single mass. @Drongoski will prob be better to ask coz i haven’t done physics in a while

 

[kkk579]

Another q i had is that would fricition oppose the direction innate movement or direction of applied movement? for example (pic below) would the friction oppose the direction of the force that the block is getting pushed by(42.2N) or would it oppose the direction of the block sliding down the incline(which is what the block has tendency to do)?

 

[kkk579]

Also for this q the acceleration is it downwards or upwards? Because when i use either sign conventions (up + down - or down + up -) i ended up getting either -5.194 or 5.914 ms^-2 which means its accelerating downwards however when i searched it up if the apparent weight is the larger than its actual weight then its accelersting upwards

 

[liamkk112]

Also for this q the acceleration is it downwards or upwards? Because when i use either sign conventions (up + down - or down + up -) i ended up getting either -5.194 or 5.914 ms^-2 which means its accelerating downwards however when i searched it up if the apparent weight is the larger than its actual weight then its accelersting upwards View attachment 43005

it’s accelerating upwards bc of newtons 3rd law. when the lift is accelerating upwards, the person in the lift is experiencing a force upwards, which means they will push back down on the lift (and hence scales)

 

[kkk579]

it’s accelerating upwards bc of newtons 3rd law. when the lift is accelerating upwards, the person in the lift is experiencing a force upwards, which means they will push back down on the lift (and hence scales)

I keep gettibg an acceleration downwards but my magnitude is right

 

[liamkk112]

I keep gettibg an acceleration downwards but my magnitude is right

when doing the calculations u might get acceleration downwards (because you are really calculating the reaction force of the person on the lift) but u then have to justify that the force of the lift is causing the person to experience a force upwards, and the person thus pushes down on the lift due to newtons 3rd law increasing the reading on the scales, so the lift must be accelerating upwards

 

[kkk579]

bump

 

[kkk579]

when doing the calculations u might get acceleration downwards (because you are really calculating the reaction force of the person on the lift) but u then have to justify that the force of the lift is causing the person to experience a force upwards, and the person thus pushes down on the lift due to newtons 3rd law increasing the reading on the scales, so the lift must be accelerating upwards

Hmm im still confused i think ill have to watch a vid on how these elevator qs work since it wasnt rlly explicitly covered in tutor