Find the fixing force in the cable AB and the reaction force at the support point O M = 18kg

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Explanation:

The expression of the weight of the bar is given by,

W=mg

Here, g is the acceleration due to gravity

Substitute all the values

W=(18kg)(9.81m/s2)=176.58N

The expression of the angle θ

by using free body diagram is given by,

tanθ=ACBO+OC

Substitute all the values

tanθ=(1.3m)(1.2m)+(0.75m)θ=tan−1(1.31.95)θ=33.69∘

Take moment about point O and obtain the value of tension T.

∑MO=0Tsinθ×AC+W×(AO2cosθ1)−Tcosθ×CO=0

Substitute all the values

Tsin(33.69∘)×(0.75m)+(176.58N)×(1.5m2cos60∘)−Tcos(33.69∘)×(1.3m)=0T×0.416m+66.22N⋅m−T×1.081m=0−T×0.665m+66.22N⋅m=0T=−66.22N⋅m−0.665mT=99.58N

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explanation: