Question

A basketball player applies a force F = 65 lb to the rim at A. Determine the equivalent force couple system at point B, which is at the center of the rim mounting bracket on the backboard

Answer 1

from 5-12-13 triangle:

$F_z=12F/13=12\cdot65/13=60$ lb

$F_{xy}=5F/13=5\cdot65/13=25$ lb

$F_x=F_{xy}sin60\degree=25sin60\degree=21.65$ lb

$F_y=F_{xy}cos60\degree=25cos60\degree=12.5$ lb

Since there is only one force acting on the rim, the resultant force will be equal to that force $\vec{F}$ , and the couple will be equal to the moment of $\vec{F}$ about point B:

$\vec{R}=\vec{F}=21.56i-12.5j-60k$ lb

$M_B=\vec{r}_{BA}\times \vec{F}$

coordinates of points A and B:

$A(-9sin30\degree,9cos30\degree,0)=A(-4.5,7.79,0)$

$B(0.-15,-4)$

$\vec{r}_{BA}=-4.5i+22.75j+4k$

$M_B=\begin{vmatrix} i & j&k \\ -4.5 & 22.79&4\\ 21.56&-12.5&-60 \end{vmatrix}=$

$=(-22.79\cdot60+4\cdot12.5)i-(4.5\cdot60-4\cdot21.56)j+(4.5\cdot12.5-22.79\cdot21.56)k=$

$=-1317.4i-183.76j-435.10k$ lb$\cdot$in