# Not too hard tension question

1. Jul 29, 2005

### rlmurra2

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????

2. Jul 29, 2005

### Nylex

What about the other component of the tension? You've got $T\cos \theta - F_{s} - mg$, why is the mg there? You also need an equals sign in that expression because it doesn't really mean anything on its own.

3. Jul 29, 2005

### rlmurra2

Ok so $T\cos \theta - F_{s} - mg$ = ma....but would be normal force be relevant?? I still dont know how to find the angle, without knowing the tension..

4. Jul 29, 2005

### Staff: Mentor

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.

5. Jul 29, 2005

### rlmurra2

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

6. Jul 29, 2005

### sniffer

Vertically:
$$Tsin\theta+N-mg=0$$
where N is the normal force by the floor upwards, so
$$N = mg-Tsin\theta$$
horizontally:
$$Tcos\theta - \mu N = 0$$
substitute for N from the first equation, express T in terms of theta,
$$T=\frac{\mu mg}{cos\theta + \mu sin \theta}$$
minimum T is when $\frac {dT}{d\theta}=0$
So,
$$\frac{dT}{d\theta}=\frac{-\mu mg (-sin\theta + \mu cos\theta)} {(cos\theta + \mu sin\theta)^2}$$
For this to be zero, the numerator must be zero,
$$-sin\theta+\mu cos\theta = 0$$
giving angle $\theta=19.3^o$

please tell me if this is wrong.

7. Jul 29, 2005

### rlmurra2

Fx=Tcos(x)-fs=max Tcos(x)=max+fs -> T=max+fs/cos(x)
Fy=n-mg=may n=may-mg = m(ay - g)

?????????

8. Jul 29, 2005

### sniffer

what do you mean?
i use limiting equilibrium assumption

9. Jul 29, 2005

### Staff: Mentor

Maybe so, but first you have to get the correct equations for equilibrium.

Sniffer did the work for you.

10. Jul 29, 2005

### rlmurra2

Sorry I posted that after I saw your post.

11. Jul 29, 2005

### Staff: Mentor

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

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

12. Jul 29, 2005

### rlmurra2

Ok Im getting this now...I just dont understand at the part where the numerator has to be 0, where does the $$\frac{dT}{d\theta}=\frac{-\mu mg (-sin\theta + \mu cos\theta)} {(cos\theta + \mu sin\theta)^2}$$
For this to be zero, the numerator must be zero,
$$-sin\theta+\mu cos\theta = 0$$ come from?

13. Jul 29, 2005

### Staff: Mentor

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.)

14. Jul 29, 2005

### rlmurra2

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.

15. Jul 29, 2005

### Staff: Mentor

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

I would not plug in numbers until the very last step.