In a lecture I heard that if we suspend two objects of different masses (and the system is accelerating) on both sides of a pulley of no resistance with a mass-less string then the tension on both sides of the string is same - this is fine till now.
To explain that the tension is same, it was...
I think I got the issue. mg cos 30 does not give the value of T1. It gives the projection as shown in the picture.
Apart from method 1 (solving for 2 equations - Horizontal and Vertical Forces) - are there any other method? Using Vectors in any other way?
Horizontal Forces are : T1 cos 60 - T2 cos 45 = 0, i.e. 0.5 (T1) - 0.7 (T2) =0
Vertical forces : T1 sin 60 + T2 sin 45 - mg = 0, i.e. 0.86 (T1) + 0.7 (T2) - 40 = 0 where m = 4 and g = 10
Adding the two equations we get 1.36 T1 = 40, so T1 = 29.4
Now, if I think T1 = mg cos 30 (30 degree...
I think I am doing something wrong in the second method...not fully getting hold of it. mg is the vector sum of T1 and T2. Rest I am stuck.
All I am trying to do is to see if I can solve without breaking individual forces in horizontal and vertical components and then follow method 1.
Thanks for your reply.
A diagram is attached. Now, let us evaluate T1. Given m = 4 Kg and g = 10 m/s^2. If we solve it the first way, we get T1 = 29 N.
But if I take the second method then T1 = mg cos 30 = 34.6 N.
Suppose an object of mass m is hung in a steady state by two strings. One string makes an angle 60 deg. at the point of contact with the object with the horizontal line. The other string makes 45 deg.
To find the tensions in these strings (T1 and T2) we take all Horizontal Forces or their...
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