A distributed load, point load (force), and couple moment act on the Y-shaped structure, as shown below. Determine the magnitude and orientation (angle from horizontal) of a single resultant force that can be used to replace the original loading and determine the location where that resultant force should be placed along section AB to have the same effect as the original loading. 120 LBS D 15° 10' 8.5' 6.5' 50 LBS/FT B 40° C 8' 2400 LB-FT A Partial answer: Force = 497.9 LBS Location: 7.38' from A
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- The point load at D is 120 lbs at 15° from the horizontal. Resolve it into horizontal (x) and vertical (y) components: - \( F_{Dx} = 120 \cos(15^\circ) \) - \( F_{Dy} = 120 \sin(15^\circ) \) Show more…
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Determine the magnitude and direction of the resultant of the three coplanar forces given below, when they act at a point. Force $A, 10 \mathrm{~N}$ acting at $45^{\circ}$ from the positive horizontal axis. Force $B, 87 \mathrm{~N}$ acting at $120^{\circ}$ from the positive horizontal axis. Force $C, 15 \mathrm{~N}$ acting at $210^{\circ}$ from the positive horizontal axis. The space diagram is shown in Fig. $22.10$. The forces may be written as complex numbers. The resultant force $$ \begin{aligned} =& f_{A}+f_{B}+f_{C} \\ =& 10 \angle 45^{\circ}+8 \angle 120^{\circ}+15 \angle 210^{\circ} \\ =& 10\left(\cos 45^{\circ}+j \sin 45^{\circ}\right)+8\left(\cos 120^{\circ}\right.\\ &\left.+j \sin 120^{\circ}\right)+15\left(\cos 210^{\circ}+j \sin 210^{\circ}\right) \\ =&(7.071+j 7.071)+(-4.00+j 6.928) \\ & \quad+(-12.99-j 7.50) \\ =&-9.919+j 6.499 \end{aligned} $$ Magnitude of resultant force $$ =\sqrt{\left[(-9.919)^{2}+(6.499)^{2}\right]}=11.86 \mathrm{~N} $$ Direction of resultant force $$ =\tan ^{-1}\left(\frac{6.499}{-9.919}\right)=146.77^{\circ} $$ (since $-9.919+j 6.499$ lies in the second quadrant). Thus force $A, f_{A}=10 \angle 45^{\circ}$, force $B, f_{B}=8 \angle 120^{\circ}$ and force $C, f_{C}=15 \angle 210^{\circ}$
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