Question: A 250-V, 4-pole, wave-wound d.c. series motor has 782 conductors on itsarmature. It has armature and series field resistance of 0.75 ohm. The motor takesa current of 40 A. Estimate its speed and gross torque developed if it has a flux per pole of 25 mWb Answer: Step 1 Mechanical Engineering homework question answer, step 1, image 1 Mechanical Engineering homework question answer, step 1, image 2 Step 2 Mechanical Engineering homework question answer, step 2, image 1 Step 3 Mechanical Engineering homework question answer, step 3, image 1

Link will be apear in 30 seconds.

Well done! you have successfully gained access to Decrypted Link.

 Question

The cable is subjected to a uniform loading of w = 60 kN/m. Determine the maximum and minimum tension in cable.

Solution

Step 1 Solution:-

Free Body Diagram

 

Step 2 Solution:-

$$\begin{gathered} y=\frac{1}{F_H}\int \left(\int wdx\right)dx \\ Integrating we get, \\ y=\frac{1}{F_H}\int wdxdx+C_1 \\ Again by integrating we get, \\ y=\frac{1}{F_H}\left(\frac{w{x}^2}{2}+C_{1}x+C_2\right)---equation 1 \\ U\sin g the boundary conditions, \\ x=0 and \frac{dy}{dx}=0 \\ {C}_1=0 \\ Now, \\ Substitutig the value of C_{1}and C_2 into the 1 st equation we get, \\ y=\frac{1}{F_H}\left(\frac{w{x}^2}{2}\right)---equation 2 \end{gathered}$$

 

Step 3 Solution:-

$$\begin{gathered} The above equation is parabolic. \\ Now, \\ x=50 m , y = 12m \\ Now, \\ from equation 2 \\ 12=\frac{1}{F_H}\left(\frac{60*{50}^2}{2}\right) \\ {F}_H=6250kN \\ now, \\ x=50m and \theta =\theta max \\ Hence, \\ {\overline{)\frac{dy}{dx}}}_{x=30m}=\tan \theta max---equation 3 \\ Now, \\ from equation 2 \\ \frac{dy}{dx}=\frac{wx}{F_H} \\ {\overline{)\frac{dy}{dx}}}_{x=30m}=\frac{60*50}{6250} \\ {\overline{)\frac{dy}{dx}}}_{x=30m}=0.48 \\ Now, from equation 3, \\ 0.48=\tan {\theta }_{max} \\ {\theta }_{max}={\tan }^{-1}0.48 \\ {\theta }_{max}=25.641 degree \end{gathered}$$

 

 

Step 4 Solution:-

$$\begin{gathered} F_H=T_{max}\cos {\theta }_{max} \\ 6250=T_{max}*\cos 25.641 \\ {T}_{max}=6932.71kN \\ T=6.93MN \\ Maximum tension in the cable is T_{max}=6.93MN \\ Minimum tension in cable exists when \theta =0^o \\ Now, \\ {F}_H=T_{min}\cos 0 \\ {T}_{min}=F_H \\ {T}_{min}=6250kN \\ {T}_{min}=6.250MN \end{gathered}$$

Step 5 Answer:-

T_max=6932.71kN

T_min=6.250MN