Question

In: Physics

An 85 kg man is climbing a 30 kg ladder that is placed against a building...

An 85 kg man is climbing a 30 kg ladder that is placed against a building at an angle of 15 degrees. The Coefficient of friction is between the ladder is and the ground is 0.2.

A) Find the force of friction.

B) What is the maximum height the man can climb before the ladder starts to slip?

Thorough solution is much appreciated!

Solutions

Expert Solution

here,

The forces acting horizontally in the arrangement would be due to friction in ladder and ground... and a reaction from the wall to the tip of the ladder...

so Ra = Friction...
Ra = (0.3)N...

N is normal component...

The normal components acting are...

the weight of ladder, weight of man...

30 * 9.81 + 85 * 9.81 = N

N = 1128.15 N

So Ra = 0.2 * 1128.15 = 225.63 N

a)

the frictional force is 225.63 N

b)

Take moment from the base of the Ladder

We do not know the distance where the man must stand to maintain equilibrium position let it be x

l is length of ladder

and l/2 is the place where center of gravity will be present

so Taking clockwise moment positive...to maintain eq equate to zero...

Ra* lsin(15) - W(ladder)l/2cos(15) - W(man) * x * cos(15) = 0

225.63 * l *sin(15) - 30 * 9.81 * (l/2)*cos(15) - 85 * 9.81 * x cos(15) = 0

x = 0.1 * l

the maximum height is 0.1 *l
l is the length of ladder

genius_generous answered 2 years ago
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