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F2 = 2.8 kNF130°

Question

The two forces shown act in the x-y plane of the T-beam cross section. If it is known that the resultant R of the two forces has a magnitude of 3.5 kN and a line of action that lies 15º above the negative x-axis, determine the magnitude of F1 and the inclination (angle, theta) of F2

F2 = 2.8 kN
F1
30°
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F2 = 2.8 kN F1 30°

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Step 1

The resultant force of the two force R = 3.5 kN

The inclination of the line of action = 15o above the negative x-axis.

Step 2

The component of forces can be represented as shown in the figure

R 3.5 kN
F 2.8 kN
15°
x
30°
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R 3.5 kN F 2.8 kN 15° x 30°

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Step 3

The forces acting on the T-beam cross-section a...

The force components in the x-direction are
R-Σ
-3.5cos 15 sin 30° - 2.8cos
Fsin 30° 3.5cos15° -2.8 cose
cose 3.38-0.5F.
2.8
...1)
The force components in the y-direction are
R -ΣΕ
3.5sin15 F cos 30° 2.8sin e
2.8sin F cos30° -3.5sin 15°
sin 0.866F0.9058
2.8
...2)
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The force components in the x-direction are R-Σ -3.5cos 15 sin 30° - 2.8cos Fsin 30° 3.5cos15° -2.8 cose cose 3.38-0.5F. 2.8 ...1) The force components in the y-direction are R -ΣΕ 3.5sin15 F cos 30° 2.8sin e 2.8sin F cos30° -3.5sin 15° sin 0.866F0.9058 2.8 ...2)

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