$=14\mathrm{.}5$

$L=1in$

$d_m=0\mathrm{.}95in$

$e=64\mathrm{.}83\%$

weight

rooses of assembly $=10\mathrm{.}7$ lbf

create a FBD for the aluminum assembly on the lead screw in static equilibrium. Assume the assembly slides on two aluminum rods on either side $($account for this friction$).$ Solve for the tension. Do not neglect fraction btw nut and lead screw $(f=0\mathrm{.}15)$

Create a FBD once again But the assumbly is accelerating in the positive $x$ divetion.

$a=\frac{59}{t^2}$;$0t120$ scends

Assume there is an applied force from the torque required to move the lead screw acting in the positive $x-$divection.

torque $T=\frac{FL}{2ae}$

Solve for $T($tension$)$

if i have forgotten any necessary parameters to solve the equation then just solve symbolically.