When a 70 kg man sits on the stool, by what percent does the length of the legs decrease? Assume, for simplicity, that the stool's legs are vertical and that each bears the same load.

The diameter of one leg of the stool is missing and it's 2cm.

Answer:

(ΔL/L) = 0.00729%

Explanation:

If the Weight of the man is W, the weight will be distributed equally on the 3 legs and so the reactions for each leg will be W/3 or F/3.

Now, Youngs modulus (Y) of douglas fir wood is about 1.3 x 10^ (10) N/m^2. Gotten from youngs modulus of common materials.

Now, weight of man is 70kg.

Now diameter of one leg is 2cm. so radius of one leg = 2/2 = 1cm = 1 x 10^ (-2) m

Area for one leg is; π (1 x 10^ (-2) m) ^2 = 3.14 x 10^ (-4) m

Now as stated earlier, the force on one leg is; F/3.

Now F = mg = 70 x 9.81 = 686.7N

So, force on one leg = 686.7/3 = 228. 9N

Now we know youngs modulus (Y) = Stress/Strain.

Stress = F/A while Strain = ΔL/L

Therefore Y = (F/A) / (ΔL/L)

And therefore, (ΔL/L) = F / (AY)

So (ΔL/L) = 228.9 / (3.14 x 10^ (-4)) x (1.3 x 10^ (10)) = 7. 29 x 10^ (-5)

When expressed in percentage, it becomes 0.00729%