Question

A 4-m uniform bar of 10-kg mass is inclined 30° with the horizontal. If it is supported vertically at its ends find the force of reaction at each support.

Answer 1

$mg\cos30°=R_{ay}+R_{by}$

$mg\sin30°=R_{ax}$

$(mgl/2)\cdot\cos30°-R_{by}\cdot l=0\to$

$R_{by}=(mg/2)\cdot\cos30°=(10\cdot9.8/2)\cdot\cos30°=42.44\ (N)$ .

$R_{ax}=mg\sin30°=49\ (N)$ .

$(mgl/2)\cdot\cos30°-R_{ay}\cdot l=0\to$

$R_{ay}=(mg/2)\cdot\cos30°=(10\cdot9.8/2)\cdot\cos30°=42.44\ (N)$

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$R_a=\sqrt{42.44^2+49^2}=64.82\ (N)$ . Answer

$R_b=42.44\ (N)$ . Answer