A plank 2.00 cm thick and 16.3 cm wide is firmly attached to the railing of a ship by clamps so that the rest of the board extends 2.00 m horizontally over the sea below. A man of mass 87.6 kg is forced to stand on the very end. If the end of the board drops by 4.00 cm because of the man's weight, find the shear modulus of the wood.
Pa
To find the shear modulus of the wood, we can use the formula:
Shear Modulus (G) = (F * L) / (A * Δx)
where:
F is the force applied to the plank by the man's weight,
L is the length of the overhanging part of the plank,
A is the cross-sectional area of the plank, and
Δx is the displacement of the end of the plank.
First, let's calculate the force (F) applied by the man's weight:
F = m * g
where:
m is the mass of the man (87.6 kg), and
g is the acceleration due to gravity (approximately 9.8 m/s^2).
F = 87.6 kg * 9.8 m/s^2
F ≈ 857.28 N
Next, let's calculate the length of the overhanging part of the plank (L) in meters:
L = 2.00 m
Now, let's calculate the cross-sectional area (A) of the plank in square meters:
A = thickness * width
A = 0.02 m * 0.163 m
A ≈ 0.00326 m^2
Finally, let's calculate the displacement (Δx) of the end of the plank in meters. The problem states that the end of the plank drops by 4.00 cm, so:
Δx = 0.04 m
Now we can substitute these values into the shear modulus formula to find the shear modulus (G):
G = (F * L) / (A * Δx)
G = (857.28 N * 2.00 m) / (0.00326 m^2 * 0.04 m)
G ≈ 131,758 Pa
Therefore, the shear modulus of the wood is approximately 131,758 Pa.