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Forces between identical perfectly elastic solid cubes... Watch

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    these questions always confuzzle me,
    ...there are 5 identical cubes each of mass m and a steady horizontal force P is applied to the end of the surface of the first cube...
    a) show a of block 5 is P/5m
    F=ma - a = F/m = P/5m.
    b) Resultant force on block 5
    F=ma. a = p/5m * m = p/5
    ci) acceleration of block 3
    same as a
    cii) resultant force of block 3
    same as b.

    Now these bits confuzzle me... :/

    d) resultant and direction for:
    i) block 2 on block 3
    ii) block 4 on block 3

    a - cii make sense and i understand ... i think, probli not properly tho as i cant do d

    Explnations gr8ly appreciated
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    Communication of physics is probably as important as the physics itself.

    I don't see why someone should take the time to explain anything to you if you don't take the miniscule amount of time required to type that extra letter here and there to make your request more readable.

    Or if you prefer,

    Coms of fizx S prolly as impt as d fizx Itslf. I dnt CY sum1 shd taK d tym 2 Xpln NEfin TU f u dnt taK d miniscule amt of tym needed 2 typ dat xtra letta hre n der 2 mke yr RQ mor readable.

    Which I know you don't, because it's hard to read.

    (inb4 too harsh)
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    these questions always confuzzle me,
    ...there are 5 identical cubes each of mass m and a steady horizontal force P is applied to the end of the surface of the first cube...
    a) show a of block 5 is P/5m
    F=ma - a = F/m = P/5m.
    b) Resultant force on block 5
    F=ma. a = p/5m * m = p/5
    ci) acceleration of block 3
    same as a
    cii) resultant force of block 3
    same as b.

    Now these bits confuzzle me... :/

    d) resultant and direction for:
    i) block 2 on block 3
    ii) block 4 on block 3

    a - cii make sense and i understand ... i think, probli not properly tho as i cant do d

    Explnations gr8ly appreciated
    I remember it took me a while to understand this

    For d).i.
    3P/5 resultant for 2 on 3. Same direction as travel.Think of blocks 3,4 and 5 as one block of 3m, accelerating at 'a'. Force on first cube P=5ma(to all 5 to move at 'a'), here 3/5 of total mass, use Newton's Second Law.

    d).ii.
    Block 4 on 3 is acting against P hence opposite, force 2P/5 Think of blocks 1,2 and 3 as one of 3m, and 4,5 one of 2m. All accelerating again at 'a'. What would be the acceleration of 1,2,3 if 4 wasn't exerting on 3? P/3m.

    With the force highlighted it is P/5m.
    Difference in acceleration due to counteracting force on 1,2,3 is P/3m-P/5m=2P/15m on mass 3m. Using F=ma gives 2P/15m*3m=2P/5 acting against the motion.

    Newtons Laws 1,2,3 should get you through any of these questions. Message me if any problems :ciao:
    Josh
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    (Original post by PurpleRedSky)
    I remember it took me a while to understand this

    For d).i.
    3P/5 resultant for 2 on 3. Same direction as travel.Think of blocks 3,4 and 5 as one block of 3m, accelerating at 'a'. Force on first cube P=5ma(to all 5 to move at 'a'), here 3/5 of total mass, use Newton's Second Law.

    d).ii.
    Block 4 on 3 is acting against P hence opposite, force 2P/5 Think of blocks 1,2 and 3 as one of 3m, and 4,5 one of 2m. All accelerating again at 'a'. What would be the acceleration of 1,2,3 if 4 wasn't exerting on 3? P/3m.

    With the force highlighted it is P/5m.
    Difference in acceleration due to counteracting force on 1,2,3 is P/3m-P/5m=2P/15m on mass 3m. Using F=ma gives 2P/15m*3m=2P/5 acting against the motion.

    Newtons Laws 1,2,3 should get you through any of these questions. Message me if any problems :ciao:
    Josh

    Thank you sooo much !

    small question though, for part dii) can you simply apply Newtons 3rd law that the magnitude of the force exerted on block 3 by 4 is equal to the magnitude of the force exerted on 4 by 3, and then simply use the same method as di gives 2p/5?

    thanks alot
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    (Original post by rainbowsss)
    Thank you sooo much !

    small question though, for part dii) can you simply apply Newtons 3rd law that the magnitude of the force exerted on block 3 by 4 is equal to the magnitude of the force exerted on 4 by 3, and then simply use the same method as di gives 2p/5?

    thanks alot
    Sorry for taking so long to reply
    Sure, doing it that way is probably easier actually, 4 and 5 accelerating at 'a' mass 2m. And yes, Newton's Third Law shows that this force is same as 4 on 3
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    (Original post by PurpleRedSky)
    I remember it took me a while to understand this

    For d).i.
    3P/5 resultant for 2 on 3. Same direction as travel.Think of blocks 3,4 and 5 as one block of 3m, accelerating at 'a'. Force on first cube P=5ma(to all 5 to move at 'a'), here 3/5 of total mass, use Newton's Second Law.

    d).ii.
    Block 4 on 3 is acting against P hence opposite, force 2P/5 Think of blocks 1,2 and 3 as one of 3m, and 4,5 one of 2m. All accelerating again at 'a'. What would be the acceleration of 1,2,3 if 4 wasn't exerting on 3? P/3m.

    With the force highlighted it is P/5m.
    Difference in acceleration due to counteracting force on 1,2,3 is P/3m-P/5m=2P/15m on mass 3m. Using F=ma gives 2P/15m*3m=2P/5 acting against the motion.

    Newtons Laws 1,2,3 should get you through any of these questions. Message me if any problems :ciao:
    Josh
    Hi josh, i really don't follow what you just wrote, its my bad i'm having a blond day...
    do you think you could break it down a bit more for me please mate
 
 
 
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