# M1 - MomentsWatch

#1
Tried doing this, even made friends doing fm try do it, but know onecould do it.
Someone help me do this

A uniform rod PQ of mass 4kgand length 3m is suspended from the ceiling by two strings attached to the two ends and the rod is horizontal. A particle of mass (m) kg is attached to a rod at R where PR= 0.5m. The tension in the string at end P is twice the tension in the string at end Q. Find the value of m.
0
9 years ago
#2
Well you know that the forces going upwards equal forces going downwards, then you just sub what T is into the moments equation.

I got the mass as 3.7 Kg
0
#3
(Original post by djpailo)
Well you know that the forces going upwards equal forces going downwards, then you just sub what T is into the moments equation.

I got the mass as 3.7 Kg
Could you explain it more briefly please.
0
9 years ago
#4
you should draw the diagram for it 1st and then it will more clearly to solve the problem
0
9 years ago
#5
Okay, so take moments about the centre of the rod. Since its uniform, all the weight acts towards the centre. Then, since the rod is 3m in length, you can work out an equation that equals zero, since the rod is in equilibrium. Remember that tension is already a force so you don't need to multiply the tension with g. The tension acts upwards.

When you have derived that equation, you will have three variables, g (which is 9.8), m and t the unknowns.

All the upwards forces equal the downward forces. Therefore you can derive a second equation just relating all the forces on the rod.

Now you have two simultaneous equations and two unknown variables, so you sub everything back in.
0
#6
(Original post by djpailo)
Then, since the rod is 3m in length, you can work out an equation that equals zero, since the rod is in equilibrium. Remember that tension is already a force so you don't need to multiply the tension with g. The tension acts upwards.
how would you do this?
i dont seem to get the right answer...its 4 at the back ofthe book..but im getting 0.4

Any help on forming equations??
0
9 years ago
#7
Post your working, and someone will be able to check it.
#8
ive figured it out! thanks
0
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