Calculate Moment: 2.5kN Force about Point C

[salman213]

1. It is known that the connecting rod AB exerts on the crank BC a 2.5kN force directed down and to the left along the center line of AB. Determine the moment of that force about C.

http://img267.imageshack.us/img267/6822/fic03p013hb0.png



2. cross product



3.
I found the angle using tan-1 = 144/42 = 73.7

Fx = -2.5cos73.7 = -0.702
Fy = -2.5sin73.7 = -2.40

F = (-0.702i -2.40j + 0k)
CB = (-42i -56j + 0k)


F x CB= (-0.702i -2.40j + 0k) x (-42i -56j + 0k)
= -61.5 kN mm

does that seem right??

 

[salman213]

that first was the vector approach but a scalar approach is


y and x compents then multiply by distance and just add them

Fx = 2.5 x cos 73.7 = 0.702
Fy = 2.5 x sin 73.7 = 2.40

Clockwise = -
Counterclockwise = +

Mx = -0.702 x 56
My = 2.40 x 42

M = Mx + My = 61.5 kN mm


can someone make sure I am interpreting the question accurately ! :)

thanks!

 

[salman213]

Anyone pleasezzzzzzz

 

[salman213]

no knows?

 

[Astronuc]

I found the angle using tan-1 = 144/42 = 73.7

Fx = -2.5cos73.7 = -0.702
Fy = -2.5sin73.7 = -2.40

I don't have a lot of time at the moment, but 73.7° is the angle of arm AB with respect to horizontal. Are you sure. The 2.5 kN acts through AB at angle of 16.3° with respect to vertical.

To find the moment on BC, one wants the force component from AB normal to BC.

 

[salman213]

hmmm huh? so what i did was wrong?...

even if i use 16.3 basically the y component of the force AB is going to be

-2.5cos16.3= which is what i found -2.40
and x component
-2.5sin16.3= -0.702

these are the components of AB if i want moment about C i can take a vector from C to B and cross it with AB right?

AB x CB = 61.5 kN mm

 

[salman213]

http://img229.imageshack.us/img229/4462/47859852fi8.jpg

 

[salman213]

can someone atleast explain properly what I am doing wrong if I am doing anything wrong...