It is known that the connecting rod AB exerts on the crank BCa 2.5-kN force directed down andto the left along the centerline ofAB. Determine the moment of this force about C for the two casesshown.

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shahnoorazhar3 shahnoorazhar3 · Answer 1 · 2020-02-05T14:07:27+01:00

Answer:

M_c = 61.6 Nm

Explanation:

Given:

F_a = 2.5 KN

Find:

Determine the moment of this force about C for the two casesshown.

Solution:

- Draw horizontal and vertical vectors at point A.

- Take moments about point C as follows:

M_c = F_a*( 42 / 150 ) *88

M_c = 2.5*( 42 / 150 ) *88

M_c = 61.6 Nm

- We see that the vertical component of force at point A passes through C.

Hence, its moment about C is zero.

Cricetus Cricetus · Answer 2 · 2021-11-25T19:27:06+01:00

The moment of thus force will be "61.6 Nm".

Given:

Force,

Now, according to the question,

The moment of the force will be:

→ [tex]M_c = F(\frac{42}{150} )\times 88[/tex]

[tex]= 2.5\times (\frac{42}{`150} )\times 88[/tex]

[tex]= 61.6 \ Nm[/tex]

If the vertical components force at a point (point A) passes through point C, the moment about C will be "0" zero.

Thus the answer above is right.

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