Answer to Question #298240 in Trigonometry for Harsh

Question

A 70 lb bag of rice is being pulled by a person by applying a force F at an angle θ as shown. The force required to drag the bag is given by: F(θ)=70μ/μsinθ+cosθ where μ=0.35 is the friction coefficient. (a) Determine F(θ) for θ= 5°, 10°, 15°, 20°, 25°, 30°, 35° (b) Determine the angle θ where F is minimum. Do it by creating a vector θ with elements ranging from 5°to 35°and spacing of 0.01. CalculateF for each value of θ and then find the maximum F and associated θ with MATLAB's built-in function max

Answer 1

Solution.

$th=5:5:35; mu=0.35;\newline a) \newline F=70*mu./(mu*sind(th)+cosd(th)) \newline b) \newline thb=5:0.01:35; \newline Fb=70*mu./(mu*sind(thb)+cosd(thb)); \newline [Fm, i]=min(Fb);\newline Fmin=Fb(i)\newline atTheta=thb(i)\newline Command \space Window:\newline F = 23.86\newline 23.43 \newline 23.19 \newline 23.13 \newline 23.24 \newline 23.53 \newline 24.02\newline Fmin = 23.12\newline atTheta = 19.29 \newline Answer: \newline F = 23.86\newline 23.43 \newline 23.19 \newline 23.13 \newline 23.24 \newline 23.53 \newline 24.02\newline Fmin = 23.12\newline atTheta = 19.29$