Physics, 17.12.2019 07:31 harleycochran2ovyt3n

Aladder of length l is placed against a smooth wall such that it forms a angle θ with the wall and the friction force between the ladder and the floor is fs .

a. on your paper, derive an expression for the ladder's mass, m, in terms of θ, g, fs, and/or l.

b. try out your derivation. if θ=32° and the friction force on the floor is 47 n, what is the mass of the ladder?

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Answer from: washingtonisaia

a. $m=\frac{2f_{s}tan\theta}{g}$

b. m=5.99kg

Explanation:

In order to solve this problem, we can start by drawing a diagram of the situation. Drawing a diagram is really important since it will help use understand the problem better and analyze it as well. (See attached picture).

In the diagram we can see the forces that are acting on the ladder. We will assume the ladder is static (this is it doesn't have any movement) and analyze the respective forces in the x-direction and the forces in the y-direction, as well as the moments about point B.

So we start with the sum of forces about y, so we get:

$\sum F_{y}=0$

N-W=0

N=W

N=mg

Next we can do the sum of forces about x, so we get:

$\sum F_{x}=0$

which yields:

$f_{s}-F_{W}=0$

so:

$f_{s}=F_{W}$

Next the torque about point B, so we get:

$\sum M_{B}=0$

so:

$f_{s} L sin\theta - NLcos\theta + \frac{WL}{2}cos\theta = 0$

From the sum of forces in the y-direction we know that N=mg (this is because the wall makes no friction over the ladder) so we can directly substitute that into our equation, so we get:

$f_{s} L sin\theta - WLcos\theta + \frac{WL}{2}cos\theta = 0$

We can now combine like terms, so we get:

$f_{s} L sin\theta -\frac{WL}{2}cos\theta = 0$

we know that W=mg, so we can substitute that into our equation, so we get:

$f_{s} L sin\theta -\frac{mgL}{2}cos\theta = 0$

which can now be solved for the mass m:

$\frac{mgL}{2}cos\theta = f_{s} L sin\theta$

If we divided both sides of the equation into L, we can see that the L's get cancelled, so our equation simplifies to:

$\frac{mg}{2}cos\theta = f_{s}sin\theta$

we can now divide both sides of the equation into g so we get:

$\frac{m}{2}cos\theta = \frac{f_{s}sin\theta}{g}$

next we can divide both sides of the equation into cos θ so we get:

$\frac{m}{2}= \frac{f_{s}sin\theta}{g cos\theta}$

and finally we can multiply both sides of the equation by 2 so we get:

$m=\frac{2f_{s}sin\theta}{g cos\theta}$

we know that:

$tan \theta=\frac{sin \theta}{cos \theta}$

so we can simplify the equation a little more, so we get:

$m=\frac{2f_{s}tan \theta}{g}$

So now we can directly use the equation to find the mass of the ladder with the data indicated by the problem:

θ=32° and $f_{s}=47N$

we also know that $g=9.8m/s^{2}$

so we can use our equation now:

$m=\frac{2f_{s}tan \theta}{g}$

so we get:

$m=\frac{2(47N)tan (32^{o})}{9.8m/s^{2}}$

which yields:

m=5.99kg

Aladder of length l is placed against a smooth wall such that it forms a angle θ with the wall and t

Answer from: Quest

what was the answer?

Answer from: Quest

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