Understanding Torque: The Role of Normal Force in Stability

[nomorevishnu]

IIT question...help me...

hi all...

I have a block in my hand...i am pressing it against my bedroom wall
i have applied enough force so that the block doesn't fall down...now...i have been asked a question based on this by the great IIT people...they asked me...which force is capable of producing a torque if i draw a free body diagram...i couldn't think about torque here at all...but the answer they gave is that the normal force is capable of producing torque here...how?
can anyone explain?

vishnu here

 

[whozum]

As far as I can see,theres no reason the normal force can produce a torque and not your hand.

 

[nishant]

this question came in this year's screening,although it was not mentioned but the force was aplied symmytrically on the cube due to which the force could not produce a torque about the centre of mass,and out of all the options the answer which uwrote was the only one relevant.

 

[xanthym]

hi all...

I have a block in my hand...i am pressing it against my bedroom wall
i have applied enough force so that the block doesn't fall down...now...i have been asked a question based on this by the great IIT people...they asked me...which force is capable of producing a torque if i draw a free body diagram...i couldn't think about torque here at all...but the answer they gave is that the normal force is capable of producing torque here...how?
can anyone explain?

<br /> \setlength{\unitlength}{0.002cm}<br /> \begin{picture}(6000,6000)(1500,-6000)<br /> \thinlines<br /> \linethickness{1pt}<br /> \color{red}<br /> \put(1801,-361){\line( 1, 0){2100}}<br /> \put(3901,-361){\line( 0,-1){5400}}<br /> \put(3901,-5761){\line(-1, 0){2100}}<br /> \put(3901,-1561){\line( 1, 0){1800}}<br /> \put(5701,-1561){\line( 0,-1){3000}}<br /> \put(5701,-4561){\line(-1, 0){1800}}<br /> \put(4801,-3061){\vector( 1, 1){900}}<br /> \put(7201,-2161){\vector(-1, 0){1500}}<br /> \put(6001,-2011){\line(-2,-1){300}}<br /> \put(5701,-2161){\line( 2,-1){300}}<br /> \put(5701,-2161){\line(-2,-1){300}}<br /> \put(5701,-2161){\line(-1,-2){150}}<br /> \put(4930,-2600){$\color{blue}\vec{R}$}<br /> \put(6451,-2050){$\color{blue}\vec{F}$}<br /> \put(4801,-3361){P}<br /> \put(5400,-2050){Q}<br /> \put(7351,-2250){HAND}<br /> \put(6800,-3100){\color{blue}\textsf{\LARGE Total NET Torque = 0}}<br /> \put(6800,-3450){\color{blue}\Large However, HAND force F applies torque \, $ \vec{\tau}_{ \left ( F \ about \ P \right ) } $ about point P.} <br /> \put(7351,-1600){\color{blue}$ \vec{\tau}_{ \left ( F \ about \ P \right ) } \ = \ \vec{R}\times\vec{F} \ \neq \ 0 $}<br /> \put(4400,-1411){BRICK}<br /> \put(2401,-2161){WALL}<br /> \put(1500,-6000){.}<br /> \end{picture}<br />

Of course, the Total NET Torque is ZERO (0). However, any individual force can apply a torque about a particular point. Shown above is Point "P" near the Brick's center about which HAND Force "F" applies torque "τ_{F about P}". No matter where Force "F" is applied, a point can generally be found about which the torque from that force is non-zero.



~~

 

[nomorevishnu]

answer not clear...

well i m not able to picturise why the normal force produces torque...


vishnu...

 

[xanthym]

well i m not able to picturise why the normal force produces torque...

Because torque "τ" is mathematically determined by TWO points: 1) Point "Q" where Force "F" is applied, and 2) Reference Point "P" about which "τ" is calculated. Force "F" might be normal to the Brick's surface, but it's not normal to the vector "R" pointing from Reference Point "P" to Application Point "Q".


~~

 

[whozum]

Isnt that an unclear question?

 

[nishant]

well vishnu read my post ,you will understand?

 

[xanthym]

well i m not able to picturise why the normal force produces torque...

Torque vector \vec{\tau} is only defined when 3 items are specified: 1) Reference Point "P", 2) Application Point "Q", and 3) Force vector \vec{F}. Speaking generally about a "normal force" doesn't provide enough information to properly determine the torque vector. If you completely specify all 3 items above, you should understand the torque produced.


~~

 

[siddharth]

First of all, take the torqure about the center of the block. It was not specified in the IIT question paper about which point you should take the torque, but the answer assumes you take the torque about the center of the block.

Further the Force applied on the block by the hand is acting along the center of the block. Also there is friction between the wall and the block (Otherwise the block would have fallen).

Now the question is clear, draw the free-body diagram of the block and equate torques about the center of the block. With a little thought the answer should be obvious.

 

[Dr.Brain]

nomorevishnu and siddharth did u clear IIT screening?

I couldnot...Now hopes on BITSAT DCE and AIEEE

 

[nomorevishnu]

i just want to know this...what all can the normal force do in the present situation...explain relating normal force and possibility of toque...please

 

[siddharth]

Take torque about the center of the mass. Let the side length of the block be 'a'.
Now the torque due to friction is (a/2)*f. The torque due to gravity and the applied Force is 0. The only other force is the Normal force. We also know the net torque is zero because the object will not rotate. Therefore there exists a torque due to the Normal force which opposes the torque due to friction. In fact the line of action of the Normal force will not be through the center.

 

[nishant]

in the original question{in screening} it was not mentioned that you had to take torque about the centre of mass,and may be that was the only thing that was confusing

 

[nomorevishnu]

well y is the normal force not acting through the centre of mass?

 

[nishant]

it is absolutely not necessary for the normal force to always act about the centre of mass.
By the way r u in 11th

 

[nomorevishnu]

then how is the normal force acting when the force is acting through the centre...? how does it balance the other force if it doesn't act through the same line opposite to it?or is it acting in such a way that its component cancels "F"?...doubtful yaar..please help

 

[nishant]

ncert khlolo,and see the Newton's law

 

[nomorevishnu]

well nishant

i joined this website because i wanted someone to help me...if u are going to tell me to read the book...i would just ask my senior abt a good book and obtain it...i thought for the question i ask i would get a 100%
answer in a way such that a teacher is helping me...and let that be anyone but convincing...so if i could get an explanation for my previous question about normal it would be useful yaar...