From laboratory measurements, we know that a particular spectral line formed by hydrogen appears at a wavelength of 486.1 nanome
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Answer:
a. Fab = 3000 N tension
Fac = 1500 N compression
Fbc = 3000 N compression
b. Aluminum for AB, titanium for AC and BC.
c. Fc = 4000 N
Explanation:
a. Draw free body diagrams of the pins at A and B, assuming that all internal forces are in tension (pulling away from the pin).
Sum of the forces on A in the x direction:
∑F = ma
H − Fab sin 60° = 0
Fab = H / sin 60°
Fab = 2600 / sin 60°
Fab = 3000 N
Sum of the forces on A in the y direction:
∑F = ma
-Fab cos 60° − Fac = 0
Fac = -Fab cos 60°
Fac = -H / tan 60°
Fac = -2600 / tan 60°
Fac = -1500 N (it's negative, so it's actually in compression).
Sum of the forces on B in the x direction:
∑F = ma
Fab cos 30° + Fbc cos 30° = 0
Fbc = -Fab
Fbc = -H / sin 60°
Fbc = -3000 N (it's negative, so it's in compression)
b. AB is in tension, so it should be aluminum.
AC and BC are in compression, so they should be titanium.
c. Draw a free body diagram of the entire triangle ABC. Assume there is both a horizontal and vertical reaction force at C, and assume they point in the +x and +y directions.
Sum of the forces in the x direction:
∑F = ma
Fcx + H = 0
Fcx = -H
Fcx = -2600 N
Sum of the forces in the y direction:
∑F = ma
Fcy − V = 0
Fcy = V
Fcy = 3000 N
The magnitude of the net force is:
Fc² = Fcx² + Fcy²
Fc² = (-2600 N)² + (3000 N)²
Fc = 4000 N
