A 1.30 kg toaster is not plugged in. The coefficient of static friction between the toaster and a horizontal countertop is 0.400. 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?
___________N
To answer these questions, we'll need to use the concept of static friction and the equation for tension in a rope.
(a) To find the angle above the horizontal at which you should pull the cord to minimize the cord tension, we can use the fact that the force of static friction acts in the horizontal direction and opposes the pulling force.
The formula for static friction is given by:
Fs = μs * N
Where Fs is the force of static friction, μs is the coefficient of static friction, and N is the normal force. In this case, the normal force is equal to the weight of the toaster, which is given by:
N = m * g
Where m is the mass of the toaster and g is the acceleration due to gravity.
Given:
mass of toaster (m) = 1.30 kg
coefficient of static friction (μs) = 0.400
acceleration due to gravity (g) = 9.8 m/s^2
Substituting the values into the equations, we get:
N = (1.30 kg) * (9.8 m/s^2) = 12.74 N
Now, we can set up an equation for the forces in the vertical direction:
T * sin(theta) = N
Where T is the tension and theta is the angle above the horizontal.
Substituting the values, we get:
T * sin(theta) = 12.74 N
Now, we can solve for theta.
sin(theta) = 12.74 N / T
theta = sin^(-1)(12.74 N / T)
To minimize the cord tension, we need to minimize theta. The smallest possible value for sin(theta) is 1, so to minimize theta, we need sin(theta) to be equal to 1.
Therefore, sin(theta) = 1
1 = 12.74 N / T
T = 12.74 N
Now, we can find the angle above the horizontal at which the cord should be pulled to minimize the cord tension:
theta = sin^(-1)(1) = 90°
(b) With this angle, the tension (T) must be equal to the weight of the toaster:
T = 12.74 N
Therefore, the tension must be 12.74 N.
To find the angle and tension required to make the toaster start moving, we can use the concept of static friction and the forces acting on the toaster.
(a) To determine the angle at which you should pull on the cord to minimize the cord tension, we need to find the maximum angle of static friction. The maximum angle of static friction is the angle at which the force of static friction reaches its maximum value and the object just begins to move.
To find this angle, we can use the coefficient of static friction, which relates the maximum static friction force (Ff_max) to the normal force (Fn) between the toaster and the countertop:
Ff_max = μs * Fn
In this case, the normal force is equal to the weight of the toaster:
Fn = m * g
where m is the mass of the toaster (1.30 kg) and g is the acceleration due to gravity (9.8 m/s^2).
Fn = (1.30 kg) * (9.8 m/s^2)
Fn = 12.74 N
Now, we can substitute this value into the equation for maximum static friction:
Ff_max = (0.400) * (12.74 N)
Ff_max = 5.10 N
The maximum static friction force can also be calculated as the product of the cord tension (T) and the sine of the angle between the cord and the horizontal direction (θ):
Ff_max = T * sin(θ)
Setting the two expressions for Ff_max equal to each other:
T * sin(θ) = 5.10 N
We want to minimize T, so we want sin(θ) to be as large as possible. The largest value sin(θ) can take is 1, when the angle is 90 degrees.
Therefore, the angle at which you should pull on the cord to minimize the cord tension is 90 degrees above the horizontal.
(b) With this angle, we can now calculate the tension required. Using the equation we established earlier:
T * sin(90) = 5.10 N
As sin(90) is equal to 1, we can simplify the equation:
T = 5.10 N
The tension required to make the toaster start moving, with the cord pulled at an angle of 90 degrees above the horizontal, is 5.10 N.