An 85 kg person stands on a uniform 7.1-kg ladder that is 3.9 m long, as shown.

Question

An 85 kg person stands on a uniform 7.1-kg ladder that is 3.9 m long, as shown. The floor is rough; hence it exerts both a normal force, , and a frictional force, 2, on the ladder f1 903.501 f2 164.9769 The wall, on the other hand, is frictionless; it exerts only a normal force, fs Using the dimensions in the figure, find the magnitudes of fi, f2, and fs The wall, on the other hand, is frictionless; it exerts only a normal force, f3. Using the dimensions in the figure, find the magnitudes of f1, f2, and f3 Your response is within 10% of the correct value. This may be due to roundoff error, or you could have a mistake in your calculation. Carry out all intermediate results to at least four-digit accuracy to minimize roundoff error. N f 164.9769 Your response is within 10% of the correct value. This may be due to roundoff error, or you could have a mistake in your calculation. Carry out all intermediate results to at least four-digit accuracy to minimize roundoff error. N a=3.8 m mg h 0.70 m

Explanation / Answer

f1 is equal to the amount of force the floor is applying to the ladder, which is the combined weight of the both the ladder and the person. therefore :
F1 = 85kg * 9.81(gravity) + 7.1kg * 9.81(gravity) = 902.5 N


Now as you can see the system is in static equilibrium. SO that means that F2 MUST be equal to F3(or the ladder would either fall through the wall or fall backwards).


Now we need to find the center of the ladder's gravity. Using the Pythagorean theorem we get
3.9^2 = 3.8^2 + A^2
A= 0.88 m

So for f3 we need to sum the torques.

f3= ((7.1kg)(9.81)(0.88/2) + (85kg)(9.81)(0.7)) / 3.8 = 161.66 N

As stated before f2=f3
so f2 is also equal to 161.66 N

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