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Posted

Alright, I'll change the questions:

Let coefficent = a

 

Net force on body = 3-3a=...

What is the red 3, and what is the blue 3?
Posted

Now that sorts thing out!

 

In this case I have taken a to be the coefficient of friction times g the acceleration due to gravity
That isn't the way it is defined, it wouldn't be the ratio of force to force, it would be the mass to force ratio... an acceleration... not a property of the two surfaces in contact.
  • 5 months later...
Posted

That's all well and good for the two blocks. . .

 

However everyone left the poor chicken out to dry (pun intended).

 

Here's how you work out that one.

 

Because the chicken isn't moving the net force=0

Therefore, when broken into components, the X net force is 0, as is the Y net force.

 

X:

Net Force=Force 1 + Force 2

0=T1cos 30 - T2cos 45 (side note: the cosines were introduced due to the x component. T2 is negative because of the opposite direction.)

T2cos 45=T1cos 30

T2cos 45=T1

cos 30

 

The above is equation 1.

 

Y:

Net Force=Force 1 + Force 2 + weight

0=T1sin 30 + T2sin 45 - mg

Fortunately, we have a value for T1, noted in equation 1

mg=(T2cos 45/cos 30)sin 30 + T2sin 45

mg=T2cos 45tan 30 + T2sin 45 (side note:sin30/cos 30=tan 30)

mg=T2(cos 45tan 30 + sin 45)

(2kg)(9.8m/s^2)/[(0.7071)(0.5774) + (0.7071)=T2

19.6N/1.115=T2

18N=T2 (Rounded for significant digits)

 

As though by miraculous circumstances, we now have a value for T2, to substitute into our equation 1.

 

18Ncos 45/cos 30=T1

15N=T1

 

Therefore, the wire must be able to support at least 18 Newtons of tension for the chicken to remain there, rather than fall to its gruesome death.

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