# forces question

• Feb 20th 2010, 09:14 AM
william
forces question
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
• Feb 20th 2010, 09:34 AM
HallsofIvy
Is that "95" a typo for "45"? If not, where did you get it?
• Feb 20th 2010, 09:45 AM
william
Quote:

Originally Posted by HallsofIvy
Is that "95" a typo for "45"? If not, where did you get it?

it should actually be a 75, i made a typo i was working on a couple problems at once. i drew the trangle for the problem with the angles, and added 45 and 30 to get 75^o, its hard to exaplin without my diagram present, obviously, so i am just looking for a verification of the calculations as i cannot see what i did wrong
• Feb 20th 2010, 10:29 AM
e^(i*pi)
Quote:

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 $g$ your book uses. I am using $g = 9.81 \text {N kg}^{-1}$

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

Also these are common angles and the values should be known

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

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

Vertically

Down: $mg = 20g$

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

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

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

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

Horizontally

Left: $T_1 \cos (30)$
Right: $T_2 \cos (45)$

Since there is no horizontal movement these forces are equal

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

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

--------------------------------------------

We now have two equations and two variables

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

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

Spoiler:
$T_2 = T_1\, \sqrt{\frac{3}{2}}$

$T_1 + \sqrt{2} T_1\, \sqrt{\frac{3}{2}} = T_1 + T_1 \sqrt3 = 40g$

$T_1 = \frac{40g}{1+\sqrt3} = 148.63 \text{ N}$

$T_2 = \frac{40g}{1+\sqrt3} \cdot \sqrt{\frac{3}{2}} = 175.90 \text { N}$

My answers are relatively close to the book's.