Originally Posted by

**william** A mass of 20kg is suspended from a ceiling by two lengths of rope that make angles of 30^o and 45^o with the ceiling. Determine the tension in each of the ropes.

i drew the triangle and found out the other angles using triangle properties and ended up with

T1/sin45=T2/sin60=196N/sin95

T1=196(sin45)/sin95=139.1 N and T2=196N(sin60)/sin95=170.3N

the book shows the answers to be 143.48N and 175.73N and i can't figure out how i went wrong

Edit: turns out I get the book answers when I set g = 9.8. For future reference it would be an immense help if you could tell us what the book uses for g as some use 9.8, some 9.81

First of all we need to know what value of $\displaystyle g$ your book uses. I am using $\displaystyle g = 9.81 \text {N kg}^{-1}$

This is a problem with resolving the forces horizontally and vertically.

Also these are common angles and the values should be known

$\displaystyle \sin (30) = \frac{1}{2}$, $\displaystyle \cos (30) = \frac{\sqrt3}{2} $

$\displaystyle \sin (45) = \cos (45) = \frac{\sqrt2}{2}$

__Vertically__

Down: $\displaystyle mg = 20g$

Up: $\displaystyle T_1 \sin (30) + T_2 \sin (45)$

If the object is in equilibrium the resultant force is 0 and so these two are equal

$\displaystyle 20g = \frac{T_1}{2} + \frac{T_2 \sqrt{2}}{2} = \frac{T_1 + T_2 \sqrt2}{2}$

$\displaystyle T_1 + T_2\sqrt2 = 40g$ (eq1)

__Horizontally__

Left: $\displaystyle T_1 \cos (30)$

Right: $\displaystyle T_2 \cos (45)$

Since there is no horizontal movement these forces are equal

$\displaystyle T_1 \cos (30) = T_2 \cos (45)$

This is equal to $\displaystyle T_1\sqrt3 = T_2 \sqrt2$ (eq2)

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We now have two equations and two variables

$\displaystyle T_1 + T_2\sqrt2 = 40g$ (eq1)

$\displaystyle T_1\sqrt3 = T_2 \sqrt2$ (eq2)