Posted by **John** on Saturday, March 24, 2012 at 7:29am.

Two hooks A and B are fixed to a ceiling, where AB is 2.5 m. A small object of mass 3 kg is suspended from A and B by two light inextensible strings so that it is 2 m from A and 1.3 m from B. calculate the tension in strings

- M1, Physics -
**bobpursley**, Saturday, March 24, 2012 at 9:25am
so looking at the triangle formed by the ceiling and strings, you know three sides.

Angle A= angle from ceiling to 2 m string.

Law of cosines:

1.3^{2}=2^{2}+2.5^{2}-2*2*2.5CosA

solve for angle A. Then, do the same to get angle B,

Now draw a perpendicular frm the weight to the ceiling. You have two new triangles, with 90 degree angles at the top, and two angles at the bottom from the verticals to the strings. Label them C and D (C on the long side). Labele the distance from A to the perpendicular x, and the other horiztonal distance, 2.5-x

Your objective is to find angles C, D, and x

angle C= 90-A

angle D=90-B

Now x/2=cosA solve for x.

Finally, you are almost there to write the physics..

The sum of vertical forces is zero:

3g-Tension2*sinA-Tension1.3*sinB=0

now the second equation:

sum of horizontal forces is zero.

Tension2*cosA-Tension1*cosB=0

two equations, two unknowns.

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