The line of action of the force is 30 cm from joint A. Therefore, the
magnitude of the bending moment is

M_{A} = F d_{A}

= (80
N) (30 cm) = 2,400 N-cm

Joint B

Joint B

The distance from joint B to the line of action of F is found from the
Pythagorean theorem:

d_{B} = (30^{2} + 45^{2})^{0.5}
= 54.08 cm

Substitute and solve for the magnitude of the moment:

M_{B} = (80 N) (54.08 cm) = 4,327
N-cm

Joint C

Joint C

Calculate the moment arm and the moment at joint C in the same manner
as for joint B:

d_{C} = (15^{2} + 45^{2})^{0.5}
= 47.43 cm

M_{C} = (80 N) (47.43 cm) = 3,795
N-cm

Joint D

Joint D and E

Calculate the moment of F at joints D and E in the same manner as for
joints B and C:

d_{D} = (15^{2} + 15^{2})^{0.5}
= 21.21 cm

M_{D} = (80 N) (21.21 cm) = 1,697
N-cm

d_{E} = (30^{2} + 15^{2})^{0.5}
= 33.54 cm

M_{E} = (80 N) (33.54 cm) = 2,683
N-cm

Joint E

Maximum Moment

The joint with the largest moment is joint B, which has a moment of 4,327
N cm, or in more common units 43.27 N-m.

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