# Thread: Vector forces on ring

1. ## Vector forces on ring

The question is:- Four given pulls and thrusts act on a ring. Find their equilibrant. The pulls and thrusts are as follows:
N = 58 - thrust
M = 105 - pull
L = 40 - thrust
K = 72 - pull
Angles are as follows:
NM - 73 deg
NL - 149 deg
NK - 194 deg

The methodology I have used is calculate the equilibrium for the pulls and thrusts separately. Then subtract the thrusts from the pulls. The answer I get is 60.36, however the answer given is 88.5. Where have I gone wrong?

2. Hello p75213

Welcome to Math Help Forum!
Originally Posted by p75213
The question is:- Four given pulls and thrusts act on a ring. Find their equilibrant. The pulls and thrusts are as follows:
N = 58 - thrust
M = 105 - pull
L = 40 - thrust
K = 72 - pull
Angles are as follows:
NM - 73 deg
NL - 149 deg
NK - 194 deg

The methodology I have used is calculate the equilibrium for the pulls and thrusts separately. Then subtract the thrusts from the pulls. The answer I get is 60.36, however the answer given is 88.5. Where have I gone wrong?
If a force $F$ acts outwards from the origin $O$ along a line that makes a direction $\theta$ measured anticlockwise from the positive direction of $Ox$, then its components along $Ox$ and $Oy$ are
$F\cos\theta$ and $F\sin\theta$, respectively
Since the angles are measured from the direction of $N$, I assume they are all in the same sense (e.g. all anticlockwise). So we can write down the components along and perpendicular to $ON$, where $O$ is the ring, as follows:

Along $ON$:
$N: -58$

$M: 105\cos73^o = 30.70$

$L: -40\cos149^o=34.29$

$K: 72\cos194^o =-69.86$
Perpendicular to $ON$:
$N: 0$

$M: 105\sin73^o=100.41$

$L: -40\sin149^o=-20.60$

$K: 72\sin194^o=-17.42$
We find the resultant force by summing these components:
Along ON: $-62.87$

Perpendicular to ON: $62.4$
The magnitude of the resultant is therefore $\sqrt{62.87^2+62.4^2}=88.58$.