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Not too hard tension question

  • Thread starter rlmurra2
  • Start date
This one's not too hard...A 1.30kg toaster is not plugged in. The coeffficient of static friction b/t the toaster and horizontal countertop is .350. To make the toaster start moving, you carelessly pull on its electric cord. a) for the cord tension to be as small as possible, you should pull at what angle above the horizontal? b) With this angle, how large must the tension be?

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????
 
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What about the other component of the tension? You've got [itex]T\cos \theta - F_{s} - mg[/itex], why is the mg there? You also need an equals sign in that expression because it doesn't really mean anything on its own.
 
Ok so [itex]T\cos \theta - F_{s} - mg[/itex] = ma....but would be normal force be relevant?? I still dont know how to find the angle, without knowing the tension..
 

Doc Al

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The angle and the tension are the two unknowns you are trying to find. Write down the conditions for equilibrium for horizontal and vertical components. Solve those two equations for the two unknowns.

As always, start by identifying each force on the object. Then find the horizontal and vertical components of each force.
 
I feel like I'm going in a big circle here.
 
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Vertically:
[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.
 
Fx=Tcos(x)-fs=max Tcos(x)=max+fs -> T=max+fs/cos(x)
Fy=n-mg=may n=may-mg = m(ay - g)

?????????
 
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what do you mean?
i use limiting equilibrium assumption
 

Doc Al

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

Sniffer did the work for you.
 
sniffer said:
what do you mean?
i use limiting equilibrium assumption
Sorry I posted that after I saw your post.
 

Doc Al

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rlmurra2 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)

?????????
Just before the system begins to move it is still in equilibrium. The static friction will be at its maximum value, thus equal to [itex]\mu N[/itex]. The acceleration is zero.

Your equation for Fy is missing the vertical component of the tension.
 
sniffer said:
Vertically:
[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.
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)}
{(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] come from?
 

Doc Al

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To find the minima (or maxima) of a function, set its first derivative equal to zero. (The first step is to write the tension as a function of angle.)
 
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) 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.
 

Doc Al

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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)
This is good. Now to find the angle that minimizes the needed tension, take the derivative as sniffer showed.

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

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