Question

if a person bends at the waist with a straight back making an angle of θ degrees with the horizontal, then the force f exerted on the back muscles can be modeled by the equation shown below, where w is the weight of the person.
complete parts (a) through (c).
f = \frac{0.6w\sin(\theta + 90^{\circ})}{\sin 12^{\circ}}
(a) calculate f when w = 130 lb and θ = 85°.
f = □ lb
(round to the nearest pound as needed.)

Explanation:

Step1: Substitute the values of \(W\) and \(\theta\) into the formula

Given \(W = 130\) lb and \(\theta=85^{\circ}\), the formula is \(F=\frac{0.6W\sin(\theta + 90^{\circ})}{\sin12^{\circ}}\).
First, calculate \(\theta + 90^{\circ}=85^{\circ}+90^{\circ}=175^{\circ}\).
Then, \(0.6W=0.6\times130 = 78\).

Step2: Calculate \(\sin(175^{\circ})\) and \(\sin(12^{\circ})\)

We know that \(\sin(175^{\circ})=\sin(180^{\circ}-5^{\circ})=\sin(5^{\circ})\approx0.0872\) and \(\sin(12^{\circ})\approx0.2079\).

Step3: Calculate \(F\)

Substitute the values into the formula \(F=\frac{78\times\sin(175^{\circ})}{\sin(12^{\circ})}\).
\(F=\frac{78\times0.0872}{0.2079}\).
\(F=\frac{6.8016}{0.2079}\approx33\)

Answer:

\(33\)