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Particles of masses 0.7kg and 0.3kg are attached to the ends of a light inextensible string which passes over a smooth pulley. The system is released from rest and the particles move vertically.
i) Find the tension in the string
ii) State the magnitude of the resultant force exerted on the pulley by the string.
Thanks in advance!!! :)
Don't you have a diagram, or more information like the angle of the slope, which particle is on this slope and which one is hanging freely, if there is friction, etc?
Oh, the question given was as above. The diagram is as attached.
Ah, ok, I wrongly thought a slope was included :o
Ok, you know that the 0.7 particle will pull the 0.3 particle, so, the 0.7 kg particle will go down, and the 0.3 particle will go up.
Using Newton's Law...
0.7g - T = 0.7a
T - 0.3g = 0.3a
The 0.7g is greater than the tension, and hence why I put it as 0.7g - T. On the 0.3 kg particle, the tension in the strins is greater than the gravitational force acting on it, hence why I put T - 0.3g.
I'll take g = 9.8 m/s^2.
Divide the equations:
Now you can continue solving for T, the tension.
I get T = 4.116 N = 4.1 N (2 sf)
The resultant force is given by:
or
I see. Thanks a lot :D! I get it up to part (i). For part (ii), the resultant force is 0.7g - T or T - 0.3g, you mean the resultant force is 2.76 or 1.16?
Oh my, sorry, I made a big mistake...
The question asks for the resultant force acting on the pulley by the two strings.
The resultant force on the 0.7 kg particle is 2.76 N and that on the 0.3 kg particle is 1.16 N.
But the resultant force by the string on the pulley is given by the vector sum of the forces on it. First, you have the string on the left hand side, pulling with a force of 4.1 N. The string at the right hand side now also pulls with a force of 4.1 N.
Together, they pull on the string with a force of 8.2 N.
It's ok ;). Thanks a lot for your help :)!