# static friction and tension

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#### himself

A 1.30 kg toaster is not plugged in. The coefficient of static friction between the toaster and a horizontal counter top is .350. To make the toaster start moving, you carelessly pull on its electrical cord. For the cord tension to be as small as possible, roughly what angle above the horizontal should you pull?

A. 0
B. 20
C. 45
d. 90

#### PhilIvey

10+ Year Member
5+ Year Member
A 1.30 kg toaster is not plugged in. The coefficient of static friction between the toaster and a horizontal counter top is .350. To make the toaster start moving, you carelessly pull on its electrical cord. For the cord tension to be as small as possible, roughly what angle above the horizontal should you pull?

A. 0
B. 20
C. 45
d. 90

When would it maximized? The answer is when the tension is 100% parallel to the ground and this angle is 0. So, A is out. If the angle were 90 degrees, you wouldn't do any work. A perpendicular component can't do work. A magnetic field can change the direction but it can't change speed. It accelerates because the direction is changing and velocity is a vector. I mention this so you get used to intertwining topics.

What is the component that actual does work, the parallel component. So, Tcos(theta) is what does the work. Given an equal T, the angle with the greatest cos will have the largest tension force to do work, thus requiring the least amount of tension. Cos is maximal at 0 and minimal at 90. So, the cos 20 is > than cos 45.

If you had an angle of 20, you could get it to go while applying say a tension of 10 and the component force would be say 9 assuming that cos 20 =.9. Let's say 9 is necessary. Well, the cosine of 45 is less, it's .7. So, in order to generate a force of 9, you would have to apply a greater force than 10 in order to get T*.7=9.

Now, I went to great lengths to show you everything. On the test you must be able to see the effects of 0 and 90. Then, you must realize when is cos or sin maximized.

On an inclined plane the acceleration is mgsin(theta). So, as you increase the angle you increase its sliding. You must develop such intuition. Good luck man. HTH.

#### Geekchick921

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10+ Year Member
Sorry, SDN isn't the place for homework help.

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