# Crankshaft is subjected to pulls F1 and F3 parallel to z axis and F2 and F4 parallel to y axis .What are bearring reactions at A and B if the pulls are each equal to F. A is in origin Distance...

Crankshaft is subjected to pulls F1 and F3 parallel to z axis and F2 and F4 parallel to y axis .

What are bearring reactions at A and B if the pulls are each equal to F.

A is in origin

Distance AB=2a+3b. Arm of F1 is a . Arm of F2=a+b. Arm of F3=a+2b. Arm of F4=a+3b.

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The total reaction in A is parallel to the total reaction in B, it is equal in magnitude with B and it is opposite in direction.

The 2 total reactions form a couple, since F1-F3, F2-F4 form couples when the forces are of equal magnitude.

We'll write the equations of equilibrium:

Sum of torques of z axis = 0

+F2(a+b)-F4(a+3b)+By(2a+3b) = 0

-Ay(2a+3b) - F2(a+2b) + F4*a = 0

Sum of torques of y axis = 0

-F1*a + F3(a+2b) - Bz(2a+3b) = 0

Az(2a+3b) + F1(a+3b) - F3(a+b) = 0

We'll consider F1=F2=F3=F4=F and we'll open the brackets and we'll have:

Fa + Fb - Fa - 3Fb + 2aBy + 3bBy = 0

We'll eliminate and combine like terms:

-2Fb + 2aBy + 3bBy = 0 (1)

- 2aAy - 3bAy - Fa - 2bF + Fa = 0

We'll eliminate and combine like terms:

- 2aAy - 3bAy - 2bF = 0 (2)

-Fa + Fa + 2bF - 2aBz -3bBz = 0

2bF - 2aBz -3bBz = 0 (3)

2aAz+3bAz + Fa+ 3bF - Fa + Fb = 0

We'll eliminate and combine like terms:

2aAz+3bAz + 4bF = 0 (4)

**Bz = 2bF/(2a+3b)**

**By = 2bF/(2a+3b)**

**Az = -2bF/(2a+3b)**

**Ay = -2bF/(2a+3b)**