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.

https://www.chegg.com/homework-help/questions-and-answers/17-compute-acceleration-body-b-tension-cord-attached--400-lb-300-lb-f-020-q69933701

ad

Incline angel of both the inclines  =  = 36.87 ^o

Let tension in cord connected to mass 300 lb be T ₁ and tnbsion in cord connected to 400 lb be T ₂

As pulleys are massless therefore by Newtons second law on left pulley

2T ₂ - T ₁ = 0

So T ₁ = 2T ₂

Let acceleleration in block A be a _A and acceleration in block B be a _B .

We assume block B is going down and block A is going up.

Friction force on block A

F _fA =  m _A g Cos  = 0.2 X 300 X 32.2 X Cos 36.87 = 1545.6 lbs

Applying Netons second law on block A

T ₁ - m _A g Sin  - F _fA = m _A a _A

So 2T ₂ - 300 X 32.2 X Sin 36.87 - 1545.6 = 300 a _A

So  2T ₂ - 7341.6 = 300 a _A

So T ₂ - 150 a _A = 3670.8 ----------------------(1)

Friction force on block B

F _fB =  m _A g Cos  = 0.2 X 400 X 32.2 X Cos 36.87 = 2060.8 lbs

Applying Netons second law on block B

T ₂ - m _B g Sin  + F _fA = m _B a _B

So T ₂ - 400 X 32.2 X Sin 36.87 + 2060.8 = 400 a _B

So T ₂ - 5667.2 = 400 a _B

So T ₂ - 400 a _B = 5667.2 ----------------------(2)

Also as cord connected with block B will move double the distance as moved by block A

a _B = 2 a _A

Putting value of a _B is equation (2)

T ₂ - 800 a _A = 5667.2 -------------------(1)

Sustracting equation (3) from (2)

650 a _A = -1996,4

So  a _A = -3.1 ft /s ²

a _B = 2a _A = - 6.2 ft /s ²

Minus sign indicates that block B will move upward.

So acceleration of block B =  6.2 ft /s ² (upward )

Putting vale of a _B in equation (2)

T ₂ - 400 X (-6.2) = 5667.2

So T ₂ = 3187.2 lb

So acceleration of block B =  6.2 ft /s ² (upward )

Tension is cord attached to B , T ₂ = 3187.2 lb