the shaft must have 4 inches of diameter
Explanation
Step 1
given that The horsepower (hp) that a shaft can safely transmit varies jointly with its speed (in revolutions per minute, rpm) and the cube of its diameter.
[tex]Power\propto diamter^3v[/tex]so
a) If a shaft of a certain material 2 inches in diameter can transmit 36 hp at 75 rpm
[tex]\begin{gathered} 36\text{ hp=\lparen2 in\rparen}^3*75\text{ rpm*k} \\ 36=600\text{ k} \\ k=\frac{36}{600}=0.06 \end{gathered}[/tex]b)what diameter must the shaft have in order to transmit 134 hp at 35 rph
[tex]\begin{gathered} Power\propto diamter^3v \\ Power=d^3*v*k \\ replace \\ 134=d^3*35*0.06 \\ 134=d^3*2.1 \\ divide\text{ both sides by 2.1} \\ \frac{134}{2.1}=\frac{d^3*2.1}{2.1} \\ 63.80=d^3 \\ cubic\text{ root in both sides} \\ \sqrt[3]{63.80}=\sqrt[3]{d^3} \\ 3.996=d \\ rounded \\ d=4\text{ inches} \end{gathered}[/tex]therefore
the shaft must have 4 inches of diameter
I hope this helps you
