# Reaction forces in simple beam involving applied moments

## The Attempt at a Solution

The problem I'm having is that I'm not sure how to work out the reaction forces because of the applied moment of 50kN.m. I know that the formulas are the sum of forces in a single direction equals 0 and the sum of moments about a point equal 0. But does the applied moment count as a force? And as a moment do I multiply it by the distance from the point I'm taking?

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tiny-tim
Homework Helper
welcome to pf!

hi nick! welcome to pf!
… does the applied moment count as a force? And as a moment do I multiply it by the distance from the point I'm taking?
an applied moment counts as a pair of equal and opposite (but not in-line) forces …

so no, it doesn't "count" as a linear force, and its moment is the same about any point

So if my working is correct is the reaction force at A=4.3kN and B=11.7kN

tiny-tim
Homework Helper
So if my working is correct is the reaction force at A=4.3kN and B=11.7kN

(what working? )

Oops left out the working.

Taking down as positive and and clockwise as positive

sum of forces = 0 = -RA + 4x4 - RB

sum of moments about A = 0 = 16x2 + 50 - RBx7

So RB = 11.7kN and RA = 4.3kN

nvn