1is the answer (M+2m)gcotθ2 is the rightward direction ?
Two identical ladders ,each of mass M and length L are resting on the rough horizontal surface as shown in figure. A block of mass m hangs from P. If the system is in equilibrium , find the magnitude and direction of frictional force ar AandB.
I m not getting the exact answer, hints will be enough
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15 Answers
1since it's symmetrical we can consider 1 ladder.
draw the FBD ...now just right the 2 force eqns and 1 torque eqn.
maybe you're making some mistake in the direction of normal force on the ladder due to the other one..
21i think N1 should be perpendicular to the ladder..
sorry the answer is
(M+m)gcotθ2
262N1 wont be perpendicular to the ladder.
i am also getting the same answer as rishab.
21N1 will be perpendicular to the rod here.. I got the answer before
But now i cant get it
the answer i have is correct
its JEE 2005 btw
21They dont show the direction of the reaction force at the joint.
In order to make their solution correct the direction (if any) must be as shown by rishabh.
But rishabh's FBD is wrong then i guess(cz his answer is wrong)
. Will any one draw a proper FBD for me?
1
subhodip is correct the normal force will be perpendicular....and i`ve shown its two components....and the mass which is hanging exerts a force of mg on the whole system....so it will exert an equal force of mg/2 on each of the ladders....this is pretty much the fbd....i think
262@nasiko- we dont need the value or direction of normal force at that point since we can calculate torque for one ladder about P for which torque due to normal force would be zero.
(just like the fiitjee soln.)
1oh my mistake ...evrything in my solution is correct..except that tension force action the the ladder will be mg/2 and not mg.
and for that normal reaction part, we know that the normal reaction on both the ladders will be same just in exactly opposite direction...
but if it is perpendicular to the length then it shud also be perpendicular to length of the other ladder..but it wont be so (as they have to be along the same line.)
so it will be horizontal on both the ladders
1yes that`s what i have done....the components of the normal force are acting along the same lines....by the way i`m getting the answer
1debosmit in your diagram N2 will be zero. there wont be any vertical component
1how do u know that the normal force is horizontal?even the vertical normal forces can act along the same lines on the two ladders....i`m still not sure about my fbd but the answer is coming....
262the normal reactions between the ladders will be horizontal! and i'm sure of that.