+Ty 2L 3 − (315 N) L 2 − (180 N)(L) = 0
physics. Not quite sure how to solve this. According to my notes you get the answer
T_y = Tsin(12) = 506N
+33661 I assume this is:
$$T_y\times 2L^3-315L^2-180L=0$$
where the 315 and the 180 are Newtons, and L is dimensionless, so Ty is in Newtons.
Rearranging, we get:
$$T_y=\frac{315}{2L}+\frac{90}{L^2}$$
However, you need to know the value of L before you can evaluate this (L would have to be about 0.605 to get Ty as 506N)
.
+33661 I assume this is:
$$T_y\times 2L^3-315L^2-180L=0$$
where the 315 and the 180 are Newtons, and L is dimensionless, so Ty is in Newtons.
Rearranging, we get:
$$T_y=\frac{315}{2L}+\frac{90}{L^2}$$
However, you need to know the value of L before you can evaluate this (L would have to be about 0.605 to get Ty as 506N)
.
