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Physics 25 Online
OpenStudy (maheshmeghwal9):

What is the force exerted by the ceiling on the pulley system? Please help:)

OpenStudy (maheshmeghwal9):

OpenStudy (maheshmeghwal9):

Please anybody help me:)

OpenStudy (experimentx):

18*9.8 N ... if the pulley is not accelerating and the load on the system has only mass.

OpenStudy (maheshmeghwal9):

but wt is the answer???????????????????? How do u get it??????????????????????

OpenStudy (maheshmeghwal9):

@Vincent-Lyon.Fr sir help plz :)

OpenStudy (maheshmeghwal9):

System is at rest:)

OpenStudy (maheshmeghwal9):

Its answer in my book is given 240 newtons.

OpenStudy (vincent-lyon.fr):

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.

OpenStudy (maheshmeghwal9):

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.

OpenStudy (maheshmeghwal9):

can u plz make me understand wt is the actual best solution?

OpenStudy (anonymous):

can i?

OpenStudy (maheshmeghwal9):

ya:)

OpenStudy (vincent-lyon.fr):

I've just checked. It works and 240 N is correct :-)

OpenStudy (maheshmeghwal9):

ok! but how he gt it ?

OpenStudy (maheshmeghwal9):

Would u plz tell me?

OpenStudy (anonymous):

wt. of the load = 18 * 10 = 180 N

OpenStudy (anonymous):

right?

OpenStudy (maheshmeghwal9):

ya after that?

OpenStudy (anonymous):

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

OpenStudy (maheshmeghwal9):

ok then

OpenStudy (anonymous):

now, one rope at the right side is ballenced by the force F right?

OpenStudy (maheshmeghwal9):

ok

OpenStudy (anonymous):

so, F = 60 N

OpenStudy (vincent-lyon.fr):

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.

OpenStudy (anonymous):

right?

OpenStudy (vincent-lyon.fr):

Sorry, was busy typing while sauravshakya gave the same explanation.

OpenStudy (maheshmeghwal9):

np:) & both of u @sauravshakya @Vincent-Lyon.Fr sir helped me a lot Thank u very much:)

OpenStudy (maheshmeghwal9):

i gt the question. It is trickier one.

OpenStudy (vincent-lyon.fr):

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

OpenStudy (maheshmeghwal9):

ok:)

OpenStudy (experimentx):

|dw:1343478593203:dw| this is interesting ... what if the other end was fixed at rigid wall? figure not drawn accurately

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