Arescue worker pulls an injured skier lying on a toboggan (with a combined mass of 127 kg) across flat snow at a constant speed. a 2.43 m rope is attached to the toboggan at ground level, and the rescuer holds the rope taut at shoulder level. if the rescuer's shoulders are 1.65 m above the ground, and the tension in the rope is 148 n, what is the coefficient of kinetic friction between the toboggan and the snow?

0.088

Explanation:

from the question we are given the following

combined mass (m) = 127 kg

acceleration (a) = 0  (because speed is constant)

length of the rope = 2.43 m

height of the rescuers shoulder = 1.65 m

tension in the rope (T) = 148 N

acceleration due to gravity (g) = 9.8 m/s^2

coefficient of kinetic friction (μ) = ?

firstly from the diagram attached we can see that the rope and the rescuer with the ground for a right angled triangle with the rescuer being the opposite side and the rope being the hypotenuse. We can calculate the angle (θ) between the rope and the ground using the formula

sin θ = opposite / hypotenuse

θ = sin^(-1)   x   (opposite / hypotenuse ) =  sin^(-1)   x   (1.65 / 2.43 )

θ = 42.8 degrees

we can get the coefficient of kinetic friction by applying the formula below

net force = force  by the rescuer - force of kinetic friction (force as a result of kinematic friction)  .....equation 1

where

force by the rescuer = tensional force x cos θ = 148 x cos  42.8 = 100.49 Nnet force = mass x acceleration = 127 x 0 = 0 Nforce of kinetic friction = ( normal force - vertical component of tensional force ) x μ

        force of kinetic friction = (( 127 x 9.8 ) - ( 148 x sin 42.8)) x  μ

        force of kinetic friction = (1244.6 - 100.6) x μ = 1144μ

now substituting all the values in to equation 1 we have

0 =100.49 - 1144μ

1144μ = 100.49

μ = 0.088

I = 113.014 kg.m^2

m = 2075.56 kg

wf = 3.942 rad/s

Explanation:

Given:

- The constant Force applied F = 300 N

- The radius of the wheel r = 0.33 m

- The angular acceleration α = 0.876 rad / s^2

Find:

(a) What is the moment of inertia of the wheel (in kg · m2)?

(b) What is the mass (in kg) of the wheel?

(c) The wheel starts from rest and the tangential force remains constant over a time period of t= 4.50 s. What is the angular speed (in rad/s) of the wheel at the end of this time period?

Solution:

- We will apply Newton's second law for the rotational motion of the disc given by:

                                   F*r = I*α

Where, I: The moment of inertia of the cylindrical wheel.

                                   I = F*r / α

                                   I = 300*0.33 / 0.876

                                  I = 113.014 kg.m^2

- Assuming the cylindrical wheel as cylindrical disc with moment inertia given as:

                                   I = 0.5*m*r^2

                                   m = 2*I / r^2

Where, m is the mass of the wheel in kg.

                                   m = 2*113.014 / 0.33^2

                                   m = 2075.56 kg

- The initial angular velocity wi = 0, after time t sec the final angular speed wf can be determined by rotational kinematics equation 1:

                                  wf = wi + α*t

                                  wf = 0 + 0.876*(4.5)

                                  wf = 3.942 rad/s