Hello p75213

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**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 $\displaystyle F$ acts outwards from the origin $\displaystyle O$ along a line that makes a direction $\displaystyle \theta$ measured anticlockwise from the positive direction of $\displaystyle Ox$, then its components along $\displaystyle Ox$ and $\displaystyle Oy$ are$\displaystyle F\cos\theta$ and $\displaystyle F\sin\theta$, respectively

Since the angles are measured from the direction of $\displaystyle N$, I assume they are all in the same sense (e.g. all anticlockwise). So we can write down the components along and perpendicular to $\displaystyle ON$, where $\displaystyle O$ is the ring, as follows:

Along $\displaystyle ON$: $\displaystyle N: -58$

$\displaystyle M: 105\cos73^o = 30.70$

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

$\displaystyle K: 72\cos194^o =-69.86$

Perpendicular to $\displaystyle ON$:$\displaystyle N: 0$

$\displaystyle M: 105\sin73^o=100.41$

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

$\displaystyle K: 72\sin194^o=-17.42$

We find the resultant force by summing these components:Along ON: $\displaystyle -62.87$

Perpendicular to ON: $\displaystyle 62.4$

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

Grandad