Problem

Find the horsepower that the brake of Fig. 1 can absorb, and the length of arm a.Figure 1 

Find the horsepower that the brake of Fig. 1 can absorb, and the length of arm a.

Figure 1

Step-by-Step Solution

Solution 1

Sketch the free body diagram for the brake as in Figure (1).

Z:\01_Solutions_Authoring\Mechanical_Engineering\10196\04_Artwork\10196-6-9P_fig1.jpg

Refer to Figure (1):

Length of the arm is 36.6 mm.

Express friction power for the brake as:

$$ P=\frac{T n}{9,550,000} \ldots \ldots \text { (1) } $$

Here, friction power for the brake is \(P\), torque exerted by the brake is \(T\), and speed is \(n\).

To find power, calculate the torque first:

From Figure (1), torque is expressed as:

$$ T=\left(F_{1}-F_{2}\right) r \ldots \ldots \text { (2) } $$

Here, tension exerted at point 1 is \(F_{1}\), tension exerted at point 2 is \(F_{2}\), and drum radius is \(r\).

Find the torque, first calculate the tension involved.

Apply static moment of equilibrium about point \(A\).

$$ \begin{aligned} &(1350 \mathrm{~N})(300 \mathrm{~mm})+\left(F_{2}\right)(36.6 \mathrm{~mm})-\left(F_{1}\right)(75 \mathrm{~mm})=0 \\ &405,000+36.6 F_{2}-75 F_{1}=0 \end{aligned} $$

Express tensions in band brake along with angle of contact as:

$$ \frac{F_{1}}{F_{2}}=e^{j u} $$

Here, tension exerted at point 1 is \(F_{1}\), tension exerted at point 2 is \(F_{2}\), coefficient of friction is \(\mu\), and angle of contact is \(\alpha\).

Substitute \(0.15\) for \(\mu\) and \(210^{\circ}\) for \(\alpha\).

$$ \begin{aligned} &\frac{F_{1}}{F_{2}}=e^{(02)(225)\left(\frac{\pi}{180^{\circ}}\right)} \\ &F_{1}=e^{0.7854} F_{2} \\ &F_{1}=2.193 F_{2} \end{aligned} $$

Substitute Equation (4) in Equation (3) to find \(F_{2}\).

$$ \begin{aligned} &405,000+36.6 F_{2}-75\left(2.193 F_{2}\right)=0 \\ &127.875 F_{2}=405,000 \\ &F_{2}=\frac{405,000}{127.875} \\ &F_{2}=3,167 \mathrm{~N} \end{aligned} $$

Substitute \(3,167 \mathrm{~N}\) for \(F_{2}\) in Equation (4) to find \(F_{1}\).

$$ \begin{aligned} F_{1} &=2.193(3,167 \mathrm{~N}) \\ &=6,945 \mathrm{~N} \end{aligned} $$

Calculate the torque exerted by the brake using Equation (2).

Substitute 6,945 \(\mathrm{N}\) for \(F_{1}, 3,167 \mathrm{~N}\) for \(F_{2}\), and \(125 \mathrm{~mm}\) for \(r\) in Equation (2) to find \(T\).

$$ \begin{aligned} T &=(6,945 \mathrm{~N}-3,167 \mathrm{~N})(125 \mathrm{~mm}) \\ &=(3,778)(125) \\ &=472,250 \mathrm{~N} \mathrm{~mm} \end{aligned} $$

Calculate the friction power using Equation (1).

Substitute \(472,250 \mathrm{~N} \mathrm{~mm}\) for \(T\) and \(200 \mathrm{rpm}\) for \(n\) in Equation (1) to find \(P\).

$$ \begin{aligned} P &=\frac{(472,250 \mathrm{~N} \mathrm{~mm})(200 \mathrm{rpm})}{9,550,000} \\ &=9.89 \mathrm{~kW} \end{aligned} $$

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