An initially stationary box of sand is to be pulled across a floor by means of a cable in which the tension should not exceed 1100 N. The coefficient of static friction between the box and the floor is 0.35. (a) What should be the angle between the cable and the horizontal in order to pull the greatest possible amount of sand, and (b) what is the weight of the sand and box in that situation?
f_s, max = mu_s*F_n
The Attempt at a Solution
I would like some guidance on my thought process and maybe some hints on how to progress.
Defining up as +y and right as +x
Initially the box isn't moving so sum of F_x and sum of F_y is zero
From the FBD, for the y direction
F_n + T*sin(θ) - F_g = 0
so F_n = F_g - T*sin(θ), so subbing it into the static friction equation
so f_s, max = 0.35*(F_g - T*sin(θ))
= 0.35 (9.8m - 1100sin(θ)) N
From the FBD, for the x direction
T*cos(θ) - f_s = 0
T*cos(θ) = f_s
so f_s = 1100*cos(θ) N
Initially I thought θ = 0° would do, but then I thought maybe to "pull the greatest possible amount of sand", we want to find a θ value that best maximizes the cos part in the horizontal component of tension (so it overcomes the f_s, max) but at the same time best minimizes the sin part in f_s, max.
II have no idea how to do this besides maybe the derivative of f_s, max(θ) , but then I just end up with f's_max = cos(θ) = 0 and that is no help.
Maybe letting f_s = f_s,max could help, but I don't know how to deal with the pesky m or how I would solve for theta.