- #1

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Ok so I have Tcos(theta)-fs-mg

fs=force of static friction fs=uk*n = (.350)(1.30kg)(9.80m/s^2)=4.459 N

mg=12.74 N

Tcos(theta)-4.459N-12.74N ......my question is, Im not sure how to solve tension or angle without having one or the other????

- Thread starter rlmurra2
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- #1

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Ok so I have Tcos(theta)-fs-mg

fs=force of static friction fs=uk*n = (.350)(1.30kg)(9.80m/s^2)=4.459 N

mg=12.74 N

Tcos(theta)-4.459N-12.74N ......my question is, Im not sure how to solve tension or angle without having one or the other????

- #2

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- #3

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- #4

Doc Al

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As always, start by identifying each force on the object. Then find the horizontal and vertical components of each force.

- #5

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I feel like I'm going in a big circle here.

- #6

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[tex] Tsin\theta+N-mg=0[/tex]

where N is the normal force by the floor upwards, so

[tex] N = mg-Tsin\theta[/tex]

horizontally:

[tex] Tcos\theta - \mu N = 0[/tex]

substitute for N from the first equation, express T in terms of theta,

[tex] T=\frac{\mu mg}{cos\theta + \mu sin \theta}[/tex]

minimum T is when [itex]\frac {dT}{d\theta}=0[/itex]

So,

[tex]\frac{dT}{d\theta}=\frac{-\mu mg (-sin\theta + \mu cos\theta)}

{(cos\theta + \mu sin\theta)^2}

[/tex]

For this to be zero, the numerator must be zero,

[tex]-sin\theta+\mu cos\theta = 0[/tex]

giving angle [itex]\theta=19.3^o[/itex]

please tell me if this is wrong.

- #7

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Fx=Tcos(x)-fs=max Tcos(x)=max+fs -> T=max+fs/cos(x)

Fy=n-mg=may n=may-mg = m(ay - g)

?????????

Fy=n-mg=may n=may-mg = m(ay - g)

?????????

- #8

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what do you mean?

i use limiting equilibrium assumption

i use limiting equilibrium assumption

- #9

Doc Al

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Maybe so, but first you have to get the correct equations for equilibrium.rlmurra2 said:I feel like I'm going in a big circle here.

Sniffer did the work for you.

- #10

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Sorry I posted that after I saw your post.sniffer said:what do you mean?

i use limiting equilibrium assumption

- #11

Doc Al

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Justrlmurra2 said:Fx=Tcos(x)-fs=max Tcos(x)=max+fs -> T=max+fs/cos(x)

Fy=n-mg=may n=may-mg = m(ay - g)

?????????

Your equation for Fy is missing the vertical component of the tension.

- #12

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Ok Im getting this now...I just dont understand at the part where the numerator has to be 0, where does the [tex]\frac{dT}{d\theta}=\frac{-\mu mg (-sin\theta + \mu cos\theta)}sniffer said:

[tex] Tsin\theta+N-mg=0[/tex]

where N is the normal force by the floor upwards, so

[tex] N = mg-Tsin\theta[/tex]

horizontally:

[tex] Tcos\theta - \mu N = 0[/tex]

substitute for N from the first equation, express T in terms of theta,

[tex] T=\frac{\mu mg}{cos\theta + \mu sin \theta}[/tex]

minimum T is when [itex]\frac {dT}{d\theta}=0[/itex]

So,

[tex]\frac{dT}{d\theta}=\frac{-\mu mg (-sin\theta + \mu cos\theta)}

{(cos\theta + \mu sin\theta)^2}

[/tex]

For this to be zero, the numerator must be zero,please tell me if this is wrong.

[tex]-sin\theta+\mu cos\theta = 0[/tex]

giving angle [itex]\theta=19.3^o[/itex]

{(cos\theta + \mu sin\theta)^2}

[/tex]

[tex]-sin\theta+\mu cos\theta = 0[/tex] come from?

- #13

Doc Al

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- #14

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Tcos(x)-umg+uTsin(x)=0

Tcos(x)+uTsin(x)=umg

T(cos(x)+usin(x))=umg

T=umg/(cos(x)+usin(x) T=(.350)(1.30kg)(9.80m/s^2)/cos(x)+.350sin(x)

This is where Im at, I understand the part that comes next, where it comes from.

- #15

Doc Al

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This is good. Now to find the angle that minimizes the needed tension, take the derivative as sniffer showed.rlmurra2 said:Tcos(x)-u(mg-Tsin(x)=0

Tcos(x)-umg+uTsin(x)=0

Tcos(x)+uTsin(x)=umg

T(cos(x)+usin(x))=umg

T=umg/(cos(x)+usin(x)

I would not plug in numbers until the very last step.T=(.350)(1.30kg)(9.80m/s^2)/cos(x)+.350sin(x)

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