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 = 185 lb and θ = 20^{\circ}.
f = \square lb
(round to the nearest pound as needed.)

Explanation:

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

Given \(W = 185\) lb and \(\theta=20^{\circ}\), the formula is \(F=\frac{0.6W\sin(\theta + 90^{\circ})}{\sin12^{\circ}}\). First, calculate \(\sin(\theta + 90^{\circ})\). Using the trigonometric identity \(\sin(A + B)=\sin A\cos B+\cos A\sin B\), when \(A=\theta = 20^{\circ}\) and \(B = 90^{\circ}\), \(\sin(20^{\circ}+90^{\circ})=\sin110^{\circ}=\sin(90^{\circ}+ 20^{\circ})=\cos20^{\circ}\approx0.9397\).

Step2: Calculate the numerator

The numerator is \(0.6\times W\times\sin(\theta + 90^{\circ})\). Substitute \(W = 185\) and \(\sin(\theta + 90^{\circ})\approx0.9397\) into it. \(0.6\times185\times0.9397=0.6\times173.8445 = 104.3067\).

Step3: Calculate the denominator

The denominator is \(\sin12^{\circ}\approx0.2079\).

Step4: Calculate \(F\)

\(F=\frac{104.3067}{0.2079}\approx502.7\).

Answer:

\(503\) lb