Problem 6 A jungle explorer is holding a 346 lb crate on a slippery (frictionles
ID: 2304208 • Letter: P
Question
Problem 6 A jungle explorer is holding a 346 lb crate on a slippery (frictionless) 60° slone. They are using their jeep and a rope, which is wrapped part way around a rock (the angle of contact between the rope and the rock 15 60 wman n IS IT/3). The rock has a coefficient of friction of 0.387. Calculate the minimum force that the jeep must e the rope in order to maintain equilibrium. (That is to find the tension force in the rope that is between the jeep and the rock.) a) 200 lb d) 650 lb c) 500 lb b) 350 lb Hi = 0.387 W = 346lbs = 60° smooth (V2 = 0)Explanation / Answer
here,
weight of crate , W = 346 lbs
theta = 60 degree
uk = 0.387
the frictional force , ff = uk * W * cos(theta)
ff = 0.387 * 346 * cos(60) lbs
ff = 66.95 lbs
let the force applied by Jeep be F
F = ff + W * sin(theta)
F = 66.95 + 346 * sin(60) lbs
F = 350 lb
the force applied by Jeep is b) 350 lb