Question

A 68 kg crate is dragged across a floor by pulling on a rope attached to the crate and inclined 13° above the horizontal. (a) if the coefficient of static friction is 0.49, what minimum force magnitude is required from the rope to start the crate moving? (b) if μk = 0.38, what is the magnitude of the initial acceleration (m/s^2) of the crate?

Answer 1

Answer: Therefore, x component: Tcos(24°) - f = 0 y component: N + Tsin(24°) - mg = 0 The two equations I get from this are: f = Tcos(24°) N = mg - Tsin(24°) In order for the crate to move, the friction force has to be greater than the normal force multiplied by the static coefficient, so... Tcos(24°) = 0.47 * (mg - Tsin(24°)) From all that I can get the equation I need for the tension, which, after some algebraic manipulation, yields: T = (mg * static coefficient) / (cos(24°) + sin(24°) * static coefficient) Then plugging in the values... T = 283.52. Reference https://www.physicsforums.com/threads/difficulty-with-force-problems-involving-friction.111768/