Ch. 4 - Opposing Moments using Verignon's Theorem

[James Garvey]

The input fields for this exercise are linked.

[James Garvey]

[William Haynes]

James,

Thanks for reporting this.  I accidentally introduced that error a couple days ago.

Now fixed.

William Haynes Engineering Department / Professor Massachusetts Maritime Academy 100 Academy Drive Buzzards Bay, MA 02532 wha...@maritime.edu

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[James Garvey]

Found another one:


This is from Ch. 4 Exercise: Definition of a Moment. The inputs for #4 on the Moments about A are swapped.

[William Haynes]

Thanks for this one too.  I've recently been cleaning up the exercises, and a few other errors like this might have slipped in.  Keep an eye out and let me know if you see any more.

Will

[James Garvey]

I was working through this solution and wanted to double-check the moment equation and final result.

Given:
θ = 41°
A = 500

From the moment about point C:
ΣM_C = 0
B(4.4 cosθ − 3.8 sinθ) = A(6.2)

Using the values:
cos(41°) = 0.75471
sin(41°) = 0.65606

Denominator:
4.4(0.75471) = 3.3207
3.8(0.65606) = 2.4930
So:
4.4 cosθ − 3.8 sinθ = 0.8277

Moment from A:
500 × 6.2 = 3100

Solving for B:
B = 3100 / 0.8277 = 3745.328

In the provided solution, the denominator appears to use 2.49 m, which corresponds to 3.8 sin(41°) only. It seems like the 4.4 cos(θ) term may have been omitted.

Based on this, the reported value of 1081 kN does not match the full expression. I just wanted to flag this in case I am missing something in the setup or interpretation of the moment arms.

[William Haynes]

James,

Thanks for another good catch. Thanks for reporting.  I wrote and added that question last week, and there was an error in my solution.  It's all fixed now.

I appreciate your sharp eyes.

William Haynes Engineering Department / Professor Massachusetts Maritime Academy 100 Academy Drive Buzzards Bay, MA 02532 wha...@maritime.edu