QUESTION 2Kindly answer correctly. Please show the necessary steps. Thanks in advance Question 2The Figure 2 shows an internal rim-type brake having an inside rim diameter of 305 mm and adimension R=130 mm. The shoes have a face width of 40 mm and are both actuated by aforce of 2.3 kN. The drum rotates clockwise. The mean coefficient of friction is 0.28.(a) Find the maximum pressure and indicate the shoe on which it occurs.(b) Estimate the braking torque effected by each shoe, and find the total braking torque.30°-120°Pin30°-30°Pin30°120°Figure 2: Internal rim-type brake

Answered: Question 2 The Figure 2 shows an…

Elements Of Electromagnetics

7th Edition

ISBN:9780190698614

Author:Sadiku, Matthew N. O.

Publisher:Sadiku, Matthew N. O.

ChapterMA: Math Assessment

Section: Chapter Questions

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QUESTION 2

Kindly answer correctly. Please show the necessary steps. Thanks in advance

Expert Solution

Step 1

To find:

The maximum pressure and on which shoe does it acts, to estimate the braking torque on each shoe and to find the total braking force.

Given:

The inside rim radius, $r = 152.5 m m .$

The value of $R = 130 m m .$

The face width of the shoe, $b = 40 m m .$

The mean coefficient of friction, $μ = 0.28$

The force acting, $F = 2300 N .$

$θ_{1} = 0^{o}, θ_{2} = 120^{o}, θ a = 90^{o}$

Formula used:

The moment of friction force:

$\begin{array}{rcl} M_{f} & = & \frac{f \times b \times r \times P_{a}}{\sin (θ_{a})} [r - r \times \cos (θ_{2}) - \frac{R}{2} \times \sin (2 \times θ_{2})] \end{array}$

The moment of normal force

$M_{n} = \frac{P_{a} \times r \times R \times b}{\sin (θ_{a})} [\frac{θ_{2} \times π}{2 \times 180} - \frac{\sin (2 \times θ_{2})}{4}]$

The value of $C$ :

$C = 2 \times R \times \cos (30)$

The braking force for counter clockwise direction:

$\begin{array}{rcl} F & = & (\frac{M_{n} + M_{f}}{C}) \times 10^{- 4} \times P_{a} \end{array}$

The braking force for clockwise direction:

$\begin{array}{rcl} F & = & (\frac{M_{n} - M_{f}}{C}) \times 10^{- 4} \times P_{a} \end{array}$

Torque generated by right hand shoe:

$T_{R} = \frac{b \times P_{a_{m a x}} \times f \times {(r)}^{2} \times 10^{4} \times (\cos (θ_{1}) - \cos (θ_{2}))}{\sin (θ_{a})}$

Torque generated by left hand shoe:

$T_{L} = \frac{b \times P_{a_{m i n}} \times f \times {(r)}^{2} \times 10^{4} \times (\cos (θ_{1}) - \cos (θ_{2}))}{\sin (θ_{a})}$

The total torque is as follows:

$T = T_{R} + T_{L}$

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