Figure (a) shows the problem and (b) the free-body diagram. The question is to verify the reactions at C and E.

I understand the remainder of the free-body diagram but do not see how C and E are determined.

I tried force x perpendicular distance from pivot C but don't get the correct answer: (1500N x 3.5m) + (900 x 0.5) = 5700N

Any help appreciated.

enter image description here


1 Answer 1


Globally the sum of moments must be zero to be stable. You can find one of the reactions from taking moments about one of the supports (C or E), then get the rest by satisfying the equilibrium requirements $\sum F_Y = 0$ and $\sum F_X = 0$.

enter image description here

Assume roller support at C with restrain in X-direction only, and pin support at E.

Say $\sum M_E = 0$, let clockwise moment be positive, then

$F_{CX}*1.5 - 1500*5 - 900*2 = 0$

$F_{CX} = (1500*5 + 900*2)/1.5 = +6200$ - pointing to right (to produce +M as assumed).

$\sum F_Y = 0$, let the upward force be positive, thus

$F_{EY} - 1500 - 900 = 0$

$F_{EY} = 1500 + 900 = +2400$ - Acting upward.

The last, $\sum F_X = 0$, let positive force pointing to the right, so

$F_{CX} + F_{EX} = 0$

$F_{EX} = -F_{CX} = -6200$ - pointing to left.

Note that I've changed the nomenclature of the reactions to be more consistent with common practice.

  • $\begingroup$ Thank you for your response :) That makes sense and I understand where I was going wrong now. $\endgroup$
    – Racer_Rob
    Aug 24, 2021 at 8:05
  • $\begingroup$ You are welcome. $\endgroup$
    – r13
    Aug 24, 2021 at 10:56

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