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likoan [24]
3 years ago
12

A 6 m long, uniform ladder leans against a frictionless wall and makes an angle of 74.3 ◦ with the floor. The ladder has a mass

25.8 kg. A 67.08 kg man climbs 82% of the way to the top of the ladder when it slips and falls to the floor. What is the coefficient of static friction be- tween the ladder and the floor?
Physics
1 answer:
olganol [36]3 years ago
4 0

Answer: µ=0.205

Explanation:

The horizontal forces acting on the ladder are the friction(f) at the floor and the normal force (Fw) at the wall. For horizontal equilibrium,

f=Fw

The sum of the moments about the base of the ladder Is 0

ΣM = 0 = Fw*L*sin74.3º - (25.8kg*(L/2) + 67.08kg*0.82L)*cos74.3º*9.8m/s²

Note that it doesn't matter WHAT the length of the ladder is -- it cancels.

Solve this for Fw.

0= 0.9637FwL - (67.91L)2.652

Fw=180.1/0.9637

Fw=186.87N

f=186.81N

Since Fw=f

We know Fw, so we know f.

But f = µ*Fn

where Fn is the normal force at the floor --

Fn = (25.8 + 67.08)kg * 9.8m/s² =

910.22N

so

µ = f / Fn

186.81/910.22

µ= 0.205

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determine the magnitude and direction of the resultant for a combination of the vectors A: 10.1m, 60 degrees B: 7.3m, 275 degree
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5.87 m, 75.5°

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Let's say C is the resultant vector.

The x component is the sum of the x components of A and B.

Cx = Ax + Bx

Cx = 10.1 cos 60° + 7.3 cos 275°

Cx = 5.69

The y component is the sum of the y components of A and B.

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The magnitude is found with Pythagorean theorem, and the angle is found with trigonometry.

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