Mechanics Homework, Set 3
1. A worker applies a horizontal force to the top edge of a crate to get it to tip forward. If the crate has a mass of 110 KG and is 1.6 m tall and 0.80 m in depth and width, what is the minimum force needed to make the crate start tipping? (Assume the center of mass of the crate is at its center and static friction great enough to prevent slipping.) Express your answer using two significant figures.
2. A variation of Russell traction ( the figure ) supports the lower leg in a cast. Suppose that the patient leg and cast have a combined mass of 16.5 kg and m1 is 5.00 kg. A)What is the reaction force of the leg muscles to the traction? B) What must be to keep the leg horizontal?
3. While standing on a long board resting on a scaffold, a 75- painter paints the side of a house, as shown in the figure If the mass of the board is 18 kg, how close to the end can the painter stand without tipping the board over? Express your answer using two significant figures.
4. A person opens a door by applying a 14-n force perpendicular to it at a distance 0.90 m from the hinges. The door is pushed wide open (to 120 degrees ) in 2.4 s. A) How much work was done? B) What was the average power delivered? Express your answer using two significant figures.
5. An ice skater spinning with outstretched arms has an angular speed of 5.0 rad/s .She tucks in her arms, decreasing her moment of inertia by 19 percent. What is the resulting angular speed? By what factor does the skater’s kinetic energy change? (Neglect any frictional effects.)
Answers
1. The momentum of the force relatively to the point of tip must be equal to the momentum of the weight of the box relatively to the same point. let h be the height of the box and l the length (width) of the box.
$F*h = G*l/2$
$F = G*(l/2h) =110*9.81*0.8/(2*1.6 )=269.775 N$
2. A) Reaction force = 2* wire tension
$R= 2*T$
where $T = leg-mass* g + m1*g =16.5*9.81 +5*9.81 =210.915 N$
$R = 2*T = 421.83 N$
B)
The momentum of the leg mass relatively to the hip must be equal to the momentum of mass m2 relatively to the hip. Let L be the length of the leg. One must suppose that the leg weight is applied to its middle length. Then
$leg-mass*g *L/2 =m2*g*L$
$m2 =leg-mass/2 =16.5/2 =8.25 Kg$
3. The momentum of the weight of the board relatively to the point of rotation of the board must be equal to the momentum of the weight of the painter relatively to the same point. Let L be the distance from the painter to the end of the board. Let $m1$ be the mass of the painter and $m2$ the mass of the board.
$m1*(1.5-L) = m2 *2.5/2$
$1.5-L = (m2/m1) *(2.5/2)$
$1.5-L =18/75 *1.25 =0.3$
$L =1.5-0.3 =1.20 m$
4. $180 degree ……\pi (radian)$
$120 degree …..\alpha (radian)$
…………………………………………….
$\alpha =120*\pi/180 =2.09 (rad)$
the length of the displacement of the point of application of the force is
$L = \alpha*R =2.09*0.9 =1.885 m$
a) the work is
$W =F*L =14*1.885 =26.39 J$
b) the power is
$P = W/time =26.39/2.4 =10.996 W =11.00 W$
5. a) The kinetic momentum $J = I*\omega$ is the same before and after tucking the arms (I is the inertia moment, omega is the angular speed)
$I*omega_1 = (I-0.19*I)*\omega_2$
$omega_2 = 1 /(1-0.19) *\omega_1 =1/0.81 *5 =6.173 (rad/sec)$
b) the kinetic energy for a body in rotation is by definition
$Ec = I*\omega^2/2$
(I is the inertia moment, omega is the angular speed)
$Ec2/Ec1 = \omega_2^2/\omega_1^2 *I2/I1 =6.173^2/5^2 *(1-0.19)/1 =1.524*0.81 =1.234$
