Please explain, thank you. A crate, in the form of a cube with edge lengths of 1
ID: 1292196 • Letter: P
Question
Please explain, thank you.
A crate, in the form of a cube with edge lengths of 1.2 m, contains a piece of machinery; the center of mass of the crate and its contents is located 0.30 m above the crate's geometrical center. The crate rests on a ramp that makes an angle with the horizontal. As is increased from zero, an angle will be reached at which the crate will either tip over or start to slide down the ramp. If the coefficient of static friction s between ramp and crate is , (a) does the crate tip or slide and (b) at what angle does this occur? If . (c) does the crate tip or slide and (d) at what angle does this occur? (Hint: At the onset of tipping, where is the normal force located?)
Explanation / Answer
I'd find the angle at which the crate will tip. This angle is found when the weight force vector, which is drawn straight down from the center of mass, crosses the front corner of the crate.
So knowing the center of mass is at a point 0.9m from the bottom and 0.6 m from either side you can
drawn a line from that point to the front corner and determine the angle it makes with the vertical of the crate.
tan A = 0.6/0.9
A = 33.7 degrees this is the angle the crate will tip.
Now you have to find if the crate slides before 33.7 degrees
mgsinA - 0.6mgcosA = 0
tanA = 0.6
A = 30.96
So for u = 0.6 it slides at 30.96 or 31 degrees
for u = 0.7
we already know from what we've done that it has already tipped by this point.
So for u = 0.7 it tips at 33.7 degrees or I guess you rounded up to 34 degrees