Draw a diagram explaining the calculation

Assignment $2$: Eye & Vision, Muscle Contraction & Joint Rotation, Case Study Exercise

Problem $1$: The following is part of Dr$.$ Lawther's recent corrective eye glasses prescription.

Rxtor Derelk Lawther

$\backslash$table$[[,$SPHERE,CX$,$AXIS,PRISM,BASE$],[OD,-11\mathrm{.}75,-050,105,-,-],[OS,-11\mathrm{.}50,-050,095,,]]$

$($a$)$ Is the corrective prescription to correct for hyperopia or myopia? Indicate how you know this from the prescription

$($b$)$ Calculate the maximum distance from his eyes Dr$.$ Lawther can clearly see text on a piece of paper if he is not using his prescription eye glasses.

Problem $2$: An isometric exercise is a resistance technique for strengthening or rehabilitating a muscle involving muscular contractions during which no shortening or lengthening of the muscle occurs $($i$.$e$.$ no joint movement$).$

An isometric exercise for the triceps is illustrated in the diagram below.

A person places a hand on a scale and pushes vertically downward, keeping the forearm horizontal. This is possible because the triceps muscle applies an upward $($tension$)$ force M perpendicular to the forearm, as the drawing illustrates. The forearm weighs $22\mathrm{.}0$ N and has a center of mass $($cg in the diagram means center of

ANSWER KEY & HINTS

Problem $1$b:

Far$-$Points: $8\mathrm{.}51$ cm $($Right Eye$)/8\mathrm{.}70$ cm $($Left Eye$)$

You should draw a diagram explaining your calculation$/$answers

Problem $2$a:

Note that there are four forces acting on the forearm $($including the hand$).$ One of them is the upward normal force exerted by the scale in response $($Newton$\text{'}$s $3^{rd}$ law$)$ to the downward contact force the hand is exerting on the scale$).$

Problem $2$b:

The magnitude of the tension force in the contracting Triceps muscle $($M$)$ will be $1200$ N $($rounded to $2$ sig. figs.$).$ Note that if the elbow contact point on the forearm is considered as the pivot $($axis of rotation perpendicular to the page in your diagram$),$ then there are only three non$-$zero torques to consider when summing the torques and setting equal to zero $($Newton$\text{'}$s $1^{st}$ law for torques, i$.$e$.$ mechanical equilibrium$).$