What is the force exerted by the ceiling on the pulley system? Please help:)
Please anybody help me:)
18*9.8 N ... if the pulley is not accelerating and the load on the system has only mass.
but wt is the answer???????????????????? How do u get it??????????????????????
@Vincent-Lyon.Fr sir help plz :)
System is at rest:)
Its answer in my book is given 240 newtons.
I am not sure this system can work without the string slipping in the bottom pulley. Is the piece of string I pointed attached to the centre of the bottom pulley? Anyway, if the system is at rest of in steady motion, ceiling should apply an upward force of magnitude F+Weight of mass. The problem is calculating F if the system does not work! I will think about it and let you know asap.
oh ok but my sir gave solution like this but i couldn't understood it:( He said something like this that tension is same in the string.
can u plz make me understand wt is the actual best solution?
can i?
ya:)
I've just checked. It works and 240 N is correct :-)
ok! but how he gt it ?
Would u plz tell me?
wt. of the load = 18 * 10 = 180 N
right?
ya after that?
now, since this load is connected to three ropes or u can say that it is lifted by three ropes.. TENSION on each rope = 180/3 = 60N
ok then
now, one rope at the right side is ballenced by the force F right?
ok
so, F = 60 N
Your teacher gave the correct way to solve it quickly: Force F is the same along the string; eventually, bottom pulley is acted on by 4 forces: 3 equal to F pulling upwards, and 180N down. As the pulley is at rest, 3F = 180 N => F = 60 N Then whole system is at rest under reaction of ceiling R which acts upwards and force F and weight which act down. Hence R = F + 180 = 60 + 180 = 240 N My method was different, using work and velocities and I find the same result.
right?
Sorry, was busy typing while sauravshakya gave the same explanation.
np:) & both of u @sauravshakya @Vincent-Lyon.Fr sir helped me a lot Thank u very much:)
i gt the question. It is trickier one.
Note that velocity of loose end is 3 times that of the centre of the bottom pulley (v). At steady speed, power is conserved: 3vF = v x 180 when F = 60 N
ok:)
|dw:1343478593203:dw| this is interesting ... what if the other end was fixed at rigid wall? figure not drawn accurately
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