Please help! A uniform ladder of mass ( m ) and length ( L ) leans against a fri
ID: 2145458 • Letter: P
Question
Please help!
A uniform ladder of mass (m) and length (L) leans against a frictionless wall, see figure. If the coefficient of static friction between the ladder and the ground is 0.38, what is the minimum angle (q) between the ladder and the floor at which the ladder will not slip?
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A uniform ladder of mass (m) and length (L) leans against a frictionless wall, see figure. If the coefficient of static friction between the ladder and the ground is 0.38, what is the minimum angle (q) between the ladder and the floor at which the ladder will not slip?Explanation / Answer
Let's call the point where it touches the floor A and the point where it calls the wall B.
* There is the weight of the ladder located at the center of mass (in the center of the ladder), pointing down (negative y coordinate).
* The reaction force RB pushes the top of the ladder to the left. It can't have any vertical (tangential) compound, frictionless surface doesn't allow it.
* The reaction force RA pushes the bottom of the ladder to the top right. Its horizontal compound is the static friction.
Now, the sum of these three forces must be exactly zero and the same for their torques. The former gives a set of equations
RAx + RBx = 0 ==> RAx = -RBx
RAy - mg = 0 ==> RAy = mg
So there's only one unknown left.
Balancing torque, we get
2r RBx sin ? = -r mg cos ?
RBx = -cos ? / (2sin ?) mg = -1/2*cot ?*mg
now,
The x-component of the RA is 1/2*cot ?*mg, but this can't exceed the coefficient 0.38*the y-component, which is mg.
1/2*cot ?*mg < 0.38*mg
1/2*cot ? < 0.38
cot ? < 0.76
? > 52