Question

A 1.20kg toaster is not plugged in. The coefficient of static friction between the toaster and a horizontal countertop is 0.335. 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? With this angle, how large must the tension be?

Answer 1

Newton's second law allows us to find that the answers for the minimum acceleration are:

• Angle   18.5º

• Tension 3.74 N

Given parameters

• The mass of the body m = 1.20 kg

• The very rock coefficient = 0.335

To find

• Angle of stress

• The value of tension

Newton's second law says that the force is directly proportional to the masses and the acceleration of the bodies.

            F = m a

Where the bold letters indicate vectors, F is the force, m the mass and the acceleration

A reference system that allows measurements is necessary by using Newton's second law, in this case we set a reference system with the horizontal and positive x-axis towards the direction of movement and the vertical y-axis.

The free body diagram is a diagram of the forces on the body, in the attached we have a diagram of this system, let's solve for each axis

y-axis

          $T_y$ + N - W = 0

x-axis

         Tₓ - fr = m a

Where Tₓ and $T_y$ are the stress components, W is the weight of the body

(W = mg), N is the normal force and fr the friction force between the hob and the body.

The friction force is a macroscopic manifestation of the binding energy between the two surfaces, it is described by the expression

         fr = μ N

Where very is the coefficient of friction and N is the normal force

Let's use trigonometry to find the stress component

       cos θ = $\frac{T_x}{T}$Tx / T

       sin  θ = $\frac{T_y}{T}$

       Tₓ = T cos θ

       $T_y$ = T sin θ

let's substitute

       T sin θ + N - mg = 0

       T cos θ - μ N = m a

        ma = T cos θ - μ (mg - T sin θ)

        m a = T (cos θ + μ sin θ) - μ m g          (2)

To find the smallest value of the acceleration we must find the first derivative of the expression with respect to the angle

       $\frac{da}{dt } = 0$

       0 = T (- sin θ + μ cos θ ) - 0

       sin θ = μ cos θ

       tan θ = μ

       θ = tan⁻¹  μ

       θ = tan⁻¹ 0.335

       θ = 18.5º

This is the angle o that causes the acceleration to be minimal.

Let's find the stress for this angle, using equation 2

       0 = T (cos θ + μ sin θ) - μ m g

       T = $\frac{ \mu \ m g }{ cos \theta + \mu \ sin \theta}$

       T = $\frac{0.335 \ 1.20 \ 9.8 }{cos 18.5\ + 0.335 \ sin 18.5}$  

       T = $\frac{ 3.9396}{ 1.0546}$3.9396 / 1.0546

       T = 3.74 N

Using Newton's second law we can find that the answers for the minimum acceleration are:

• Angle 18.5º

• Tension 3.74 N

Learn more about Newton's second law here:

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