Physics tension of toaster quesiton?

Question

A 1.20 kg toaster is not plugged in. The coefficient of static friction between the toaster and a horizontal countertop is 0.370. To make the toaster start moving, you carelessly pull on its electric cord. For the cord tension to be as small as possible, you should pull at what angle above the horizontal?

Does anyone know how to solve this question?

A 1.20 kg toaster is not plugged in. The coefficient of static friction between the toaster and a horizontal countertop is 0.370. To make the toaster start moving, you carelessly pull on its electric cord. For the cord tension to be as small as possible, you should pull at what angle above the horizontal?

Does anyone know how to solve this question?

*how do you differentiate T = (4.3512)/(cos(theta) + 0.370sinthetha)
for theta?

Answer 1

In order for the toaster to begin to move, the force of tension pulling it in the horizontal direction must equal (in the case of minimum pulling force) the maximum force of static friction between the toaster and the counter.

If one pulls on the cord at an angle, the force of tension will be divided up into two components, one pointing in the horizontal direction (parallel to the counter and the desired motion of the toaster), and the other one pointing vertically, perpendicular to the counter. Only the horizontal component of the tension force will act to overcome the frictional force, however the vertical component (since it is acting on the opposite direction as the object’s weight) will act to reduce the normal force between the counter and the toaster and, in turn, act to reduce the maximum frictional force between the two.
Some combination of upward and horizontal force will be required to move the toaster with a minimal tension force in the cord.

The horizontal and vertical components of the tension force acting at some angle, theta, above the horizontal is given by,
F_T_x = F_T * cos (theta)
F_T_y = F_T * sin (theta)

The maximum force of static friction is:
F_f = F_n * mu = m*g*mu
Where m is the toasters mass, g is the gravitational acceleration experienced by the toaster, and mu is the coefficient of static friction between the two surfaces.
Or when taking into account that some of the tension force might be acting in the upward direction we re-write this formula as,
F_f = F_n * mu = (mg – F_T_y) * mu
Subtracting the upward tension force from the object’s weight to find the resulting normal force acting on the toaster.

In the case of minimal tension force, the horizontal component of the tension exactly equals the frictional force, so we set those two equal,

F_T * cos (theta) = (mg – F_T_y) * mu
We can solve for F_T as,
F_T = (mg*mu) / (sin (theta) * mu + cos (theta))

We are looking for what angle F_T is at its minimum. To do this, you could take the derivative of the function and find the critical points and solve for it that way. But in this case, I think there is an easier way.
Look at the function…the only place we see a variable (theta) is in the denominator. Everything besides theta is some known constant. So the function will be at its minimum when the denominator is at its maximum.
If we take the derivative of the denominator we get and setting this to zero,
mu * cost (theta) – sin (theta) = 0
Solving for mu,
mu = sin (theta) / cos (theta) = tan (theta)
so mu, the coefficient of static friction, a value we know, is equal to the tangent of the angle we must pull at.
We can solve for theta now as being the inverse tangent of mu,
Theta – arctan (mu) = tan^-1 (mu)
Theta = 20.3 degrees above the horizontal.

At this angle, the denominator function in the above tension equation is at is maximum (this tension is at is minimum).
Thus, in order to minimum the force of tension in the cord, we should pull at 20.3 degrees above the horizontal.

Answer 2

Buoyant stress is determined by technique of the formula B=PVg the position... B is the stress upwards V is the volume of the liquid displaced. g is the acceleration via gravity of the earth. Likewise, that is an same because the load of the liquid displaced. assume g become decrease than that is on earth. you would have a lot less stress pushing you upwards. notwithstanding, likewise, on the grounds that your downward stress is an same as your mass cases acceleration (i.e. gravity of the planet), and the gravity pulling you down is way less as well, both effects might want to really no longer count number. at the same time as it comes all the way down to it the in user-friendly words component that concerns is the ratio of the densities of the component you try to flow. Edit: Sorry, P contained in the unique formula become density. i'm too drained and should be interpreting for my physics very last tomorrow.

Answer 3

let angle be t whith the horizontal
then normal reaction=1.2g-fsint g is acceleration due to gravity
horizontal force=fcost
hence fcost=0.37(1.2g-fsint)
f=.37*1.2g/(cost+.37sint)
for f to be minimum denominator should be maximum diffrentiate it and find t