Answer:
Girder:
+↑ΣFy=may; 2Tsin60°−8000(9.81) = 8000 (3) T= 59166.86N ↑ΣFy=may ;\quad \quad \quad 2T \sin { 60° }- 8000(9.81) = 8000(3)\\ T = 59 166.86 N+↑ΣFy=may;2Tsin60°−8000(9.81)=8000(3)T=59166.86N
Segment:
→+ΣFx=max;59166.86cos60°−N=0N=29.6kN+↑ΣFy=may;59166.86sin60°−4000(9.81)+V=4000(3)V=0↪+ΣMC=Σ(Mk)C;M+4000(9.81)(1)–59166.86sin60°(2)=–4000(3)(1)M=51.2kN⋅m\overset { + }{ \rightarrow } \Sigma { F }_{ x }=ma_{ x };\quad \quad \quad 59166.86\cos { 60° } -N=0\\ N = 29.6 kN\\ +\uparrow \Sigma { F }_{ y }=ma_{ y };\quad \quad \quad 59 166.86 \sin { 60° }- 4000(9.81) + V = 4000(3)\\ V = 0\\ \hookrightarrow +\Sigma { M }_{ C }=\Sigma ({ M }_{ k })_{ C };\quad \quad M + 4000(9.81)(1) – 59 166.86 \sin { 60° }(2) = – 4000(3)(1)\\ M = 51.2 kN \cdot m→+ΣFx=max;59166.86cos60°−N=0N=29.6kN+↑ΣFy=may;59166.86sin60°−4000(9.81)+V=4000(3)V=0↪+ΣMC=Σ(Mk)C;M+4000(9.81)(1)–59166.86sin60°(2)=–4000(3)(1)M=51.2kN⋅m
Step-by-step explanation:
We know that Newtow Third's Law is F=ma, so we can solve for the tension on the side with that and the acceleration given in the y.
After that depending on which section of girder you chose there is a shear force and a normal force as well as a moment on the inside of it, meaning the middle of it, since is uniformly distributed, breaking down on the left the shear force is usually up, and the normal to the right, the moment counterclockwise, so from then is just a matter of doing the sumation of forces on the y and x and the summation of the moments, just remember to account for now being half of the total mass, we still have acceleration on the y, so the summation of moments will equal summation of kinetic moments, where we take into account every force that is trying to make it move, in this case, half the mass of the grinder and the acceleration going up, hope it helps you.