The Student Room Group

Mechanics ~ forces and motion in 2D

hey!

i posted a few days ago about some Q's i was having problems with and i managed to do the one Q but i'm far too stuck on the last one.

could someone have a go at doing these for me? i've tried for the past few evenings and i've got aboluslty no where :frown:

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The diagram shows a load of weight WN at B in equilibrium. The load is attached by one light string to the ceiling at A and by a second light string that passes over a smooth pulley at C to an object of mass MKG hanging freely at D. The angles of the strings and the tension T1N and 196N acting at B are shown in the diagram.

1 > i) Write down the numerical value of the tension in the string section CD, giving a reason for your answer. By considering the equilibrium of the object at D, calculate the value of m.

1 > ii) Calculate the value of T1.

1 > iii) Calculate the value of W.

1 > iv) Calculate the magnitued of the total force exerted on the pulley at C by the string passing over it.

1 > v) An addition mass, MKG, is now added at D. Explain why the system cannot be in equilibrium with ABC in a straight line no matter what the value of M.
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thank you!
where is t1n and omg this guy's room sucks!
Reply 2
we couldn't figure out where it was (never states it) so we could only presume that AB = t1n
so where is the 196
Reply 4
if AB = T1N then I presume 196N is for BC (BCD?).

this is worded so horribly!
Reply 5
I think i've done part 1 > i)

Consider that AB's tension is 196N, and as everything is in equilibrium, all forces must be equal. As its a smooth pulley, theres no friction and no break in the string, so CD's tension (BCD?!) must also be 196N (or as it only talks about CD, would you divide 196/2 so that CD's tension = 98N?). mg = 196, so m must be 20kg.

Could anyone confirm that this is right or wrong, please? :smile:

thank you!