# Equilibrium of forces... again

• Dec 30th 2008, 01:39 PM
djmccabie
Equilibrium of forces... again
Ok i finally begin to understand the resultant force questions and im hit with this one.

6. the diagram shows a body of mass 6kg suspended in equilibrium by two light inextensible strings AC and BC, attached to two fixed point A and B on the same horizontal level. The strings AC and BC are inclined at angles 45° and 60° to the horizontal respectively.

http://farm4.static.flickr.com/3103/...e50728.jpg?v=0

Find the tension in the string AC and the tension in the string BC.

Ok i really do not know how to do this.

I'm assuming you use sine and cos to do it but im not sure how at all. Could anybody at least start me off?

thanks
• Dec 30th 2008, 01:40 PM
skeeter
let $T_1$ = tension in AC

$T_2$ = tension in BC

forces up = force down ...

$T_1\sin(45) + T_2\sin(60) = 6g$

force left = force right ...

$T_1 \cos(45) = T_2 \cos(60)$

solve the system of equations for $T_1$ and $T_2$
• Dec 30th 2008, 01:55 PM
djmccabie
hi thanks, that was a lot of help. do i now find the tensions by simultaneous equations?
• Dec 30th 2008, 02:00 PM
djmccabie
so far i have T1*(√2/2) + T2*(√3/2)=58.8

T1*(√2/2) = T2*(1/2)
• Dec 30th 2008, 02:05 PM
Mush
Quote:

Originally Posted by djmccabie
Ok i finally begin to understand the resultant force questions and im hit with this one.

6. the diagram shows a body of mass 6kg suspended in equilibrium by two light inextensible strings AC and BC, attached to two fixed point A and B on the same horizontal level. The strings AC and BC are inclined at angles 45° and 60° to the horizontal respectively.

http://farm4.static.flickr.com/3103/...e50728.jpg?v=0

Find the tension in the string AC and the tension in the string BC.

Ok i really do not know how to do this.

I'm assuming you use sine and cos to do it but im not sure how at all. Could anybody at least start me off?

thanks

Let $T_{AC}$ be the tension in AC, and $T_{BC}$ is the tension in BC.

The forces can be split up into their components in the x direction, and their components in the y direction using sin and cos, indeed.

So find the x forces, and sum them, find the y forces and sum them. And then, for equilibrium, the sum of these forces is equal to zero, as such:

$F_x = T_{AC}cos(45) - T_{BC}cos(60) = 0$

$F_y = T_{AC}sin(45) + T_{BC}sin(60) -mg = 0$

Pay attention to the sign of the forces. Forces acting upwards are positive, forces acting downwards are negative. Forces acting right to left are positive, forces acting left to right are negative.

2 equations, 2 unknowns. Solve.
• Dec 30th 2008, 02:09 PM
Last_Singularity
For now, forget the value of the mass of the ball - having it be 10 kg or 100 kg does not change the reasoning that you use to solve it.

What is the ball doing? It's just hanging there, right? In other words, it is not accelerating. No accelerating implies no net force upon the ball. So find the forces and their relations. DRAW A FREE BODY DIAGRAM. ALWAYS ALWAYS ALWAYS!

Doing so, we conclude that there are three forces at play here:
1 - tension of left string: $T_L$
2 - tension of right string: $T_R$
3 - gravity acting upon the ball $F_g = mg$
They must cancel out in order to keep the ball in place.

Once again, partition the two tensions into horizontal and vertical components. So, we obtain:

Horizontal components (leftwards pull must equal rightwards pull):
1: $T_L cos 45 = T_R cos 60$

Vertical components (two strings pulling up must cancel downwards pull of gravity):
2: $T_L sin 45 + T_R sin 60 = mg$

Solve the two equations and let us know what you get (Nod)
• Dec 30th 2008, 02:20 PM
skeeter
solve for one of the tensions in terms of the other ...

T1*(√2/2) = T2*(1/2)

T1*√2 = T2

now substitute T1*√2 for T2 in the other equation and solve.
• Dec 30th 2008, 02:33 PM
djmccabie
ok so T1*(√2/2) + T1*√2(√3/2)=58.8

T1*(√2/2 + √2(√3/2)) = 58.8

T1 = (58.8) / (√2/2 + √2(√3/2))

T1 = 30.44 ??
• Dec 30th 2008, 02:43 PM
skeeter
yes
• Dec 30th 2008, 02:44 PM
djmccabie
thanks guys much appreciated!