A $720$ N bear is standing on a metal plank supported by the cable shown in the figure. At the end of the plank hangs a basket weighing $80$ N $.$ Assume the plank is uniform, weighs $200$ N $,$ is $7\mathrm{.}00$ m long, and $=60\mathrm{.}0.$

$($a$)$ When the bear is at $x=1\mathrm{.}10m,$ find the tension in the cable supporting the plank and the components of the force exerted by the wall on the left end of the plank. $($Enter the magnitudes of your answers in $N.)$

$T=$

Draw a modified force diagram, including distances. Write the equation for rotational equilibrium, with torques about the left end $of$ the plank. Solve your equation for $T.N$

$F_x=$

Apply Newton's second law $to$ the plank $in$ the horizontal direction. Solve for $F_{x^{\text{'}}}$ the horizontal force exerted $by$ the wall $on$ the plank. You will need your value for $Tto$ calculate your

answer. $N$

$F_y=$

Apply Newton's second law $to$ the plank $in$ the vertical direction. Solve for $F_y,$ the vertical force exerted $by$ the wall $on$ the plank. You will need your value for $Tto$ calculate your

answer. $N$

$($b$)$ If the cable can withstand a maximum tension of $875$ N $,$ what is the maximum distance $($in m $)$ the bear can walk before the cable breaks? $($Measure this distance from the wall.$)$

Use the torque equation you derived to calculate $T$ in part $($a$),$ and solve it for $x.$ Use the maximum value of $T.m$