All Solutions
Page 296: Lesson Check
Solve for no linear acceleration:
In order to evaluate the linear acceleration, we use the Newton’s second law:
$$
begin{align*}
F &= m ~ a \
&= m cdot 0 \
&= 0
end{align*}
$$
Therefore the total force acting on the system equals zero, if the system has no linear acceleration.
As the textbook mentions that the torque is given by the product of force and distance
$$
begin{align*}
vec{tau} &= vec{r} times vec{F} \
&= r ~ F ~ sin left( theta right) \
end{align*}
$$
Torque is a vector quantity placed in a plane perpendicular to the planes of the force and the displacement. When the acting force on the system is equal to zero, then the torque of the system is equal to zero.
Solve for the angular acceleration equals zero:
As the textbook mentions, the average angular acceleration is given by the change in angular velocity within a given time interval
$$
begin{align*}
alpha_{avg} &= dfrac{ Delta omega }{ Delta t } \
&= dfrac{ omega_{f} – omega_{i} }{Delta t } \
&= 0 \
omega_{f} &= omega_{i} \
end{align*}
$$
So, the initial angular velocity equals the final velocity.
In order to evaluate the torque, we use the following relation:
$$
begin{align*}
tau &= alpha ~ I \
&= 0 cdot I \
&= 0
end{align*}
$$
So, the torque of the system equals zero when the angular acceleration.
(b) The initial angular velocity is equal to the final angular velocity. The torque of the system equals zero when the angular acceleration.
As the textbook mentions, the balance point corresponds to zero net torque assigns that
$$
begin{align*}
tau &= m_{1} ~ g ~ r_{1} – m_{2} ~ g ~ r_{2} \
&= 0 \
m_{1} ~ r_{1} &= m_{2} ~ r_{2} \
end{align*}
$$
In general, the center of mass is the point where an object can be balanced. So, if we chose the center of mass as the point of suspension, then the torque that acting on the system is equal to zero.
Solve for the example $8.14$:
When we attach the string at the point on the croquet mallet where there equals mass on either side of the suspension point, the exerted torque due to gravity on the right piece will be greater than the exerted force on the left piece. Because the same mass is distributed farther from the suspension point on the right piece than on the left piece. So, the mallet to hang freely from the string rotates clockwise.
The force that the lady exerts on the left end is $F_{1} = 18 mathrm{~N}$ upward. The force that the man exerts on the right end of the plank is $F_{2} = 71 mathrm{~N}$ upward.
$textbf{Required: }$
Finding the weight of the child.
Solve for figure $8.19$:
As the textbook mentions, the objects in static equilibrium don’t accelerate. According to Newton’s second law:
$$
begin{align*}
sum F &= m ~ a \
&= F_{1} + F_{2} – W \
&= 0 \
end{align*}
$$
Rearrange and solve for the weight of the child:
$$
begin{align*}
W &= F_{1} + F_{2} \
&= 18 mathrm{~N} + 71 mathrm{~N} \
&= 89 mathrm{~N}
end{align*}
$$
So, the weight of the child is $89 mathrm{~N}$.
The force that the lady exerts on the left end is $F_{1} = 18 mathrm{~N}$ upward. The force that the man exerts on the right end of the plank is $F_{2} = 71 mathrm{~N}$ upward. The weight of the child is $W = 89 mathrm{~N}$. The plank’s length is $r_{p} = 2.2 mathrm{~m}$.
$textbf{Required: }$
Finding the distance from the left end of the plank to the child.
Solve for figure $8.19$:
Solve for the lady force:
As the textbook mentions that the torque is given by the product of force and distance
$$
begin{align*}
tau_{L} &= r_{L} ~ F_{L} \
&=0 mathrm{~m} times 18 mathrm{~N} \
&= 0 mathrm{~N cdot m}
end{align*}
$$
Solve for the man’s case:
As the textbook mentions that the torque is given by the product of force and distance
$$
begin{align*}
tau_{m} &= r_{p} ~ F_{m} \
&=2.2 mathrm{~m} times71 mathrm{~N} \
&= 156.2 mathrm{~N cdot m}
end{align*}
$$
Solve for the child’s case:
As the textbook mentions that the torque is given by the product of force and distance
$$
begin{align*}
tau_{c} &= r_{c} ~ W \
end{align*}
$$
Since the torque of the child is equal to the torque of the man. Rearrange and solve for the distance from the left end of the plank to the child:
$$
begin{align*}
r_{c} &= dfrac{ tau_{c} }{ W } \
&= dfrac{ tau_{m} }{ W } \
&= dfrac{ 156.2 mathrm{~N cdot m} }{ 89 mathrm{~N} } \
&= 1.755 mathrm{~m}
end{align*}
$$
So, the distance from the left end of the plank to the child is $1.755 mathrm{~m}$.
The length of the teeter-totter is $r_{t} = 2.8 mathrm{~m}$. The distance between the balance point and the left side student is $r_{1} = 1.1 mathrm{~m}$.The distance between the balance point and the right side student is $r_{2} = 1.4 mathrm{~m}$. The total mass of the two students is $m_{t} = 84 mathrm{~kg}$.
$textbf{Required: }$
Finding the mass of the left side student.
As the textbook mentions, the balance point corresponds to zero net torque assigns that
$$
begin{align*}
tau &= m_{1} ~ g ~ r_{1} – m_{2} ~ g ~ r_{2} \
&= 0 \
m_{1} ~ r_{1} &= m_{2} ~ r_{2} \
end{align*}
$$
Rearrange and solve for the value of the mass $m_{1}$:
$$
begin{align*}
m_{1} &= dfrac{ m_{2} ~ r_{2} }{ r_{1} } \
&= dfrac{m_{2} times 1.4 mathrm{~cm} }{ 1.1 mathrm{~m} } \
&= dfrac{14}{11} ~ m_{2}
end{align*}
$$
Since the total mass of the students is $m_{t} = 84 mathrm{~kg}$
$$
begin{align*}
m_{t} &= m_{1} + m_{2} \
&= m_{1} + dfrac{11}{14} ~ m_{1} \
&= dfrac{25}{14} ~ m_{1} \
&= 84 mathrm{~kg} \
end{align*}
$$
Rearrange and solve for the mass $m_{1}$:
$$
begin{align*}
m_{1} &= dfrac{14}{25} times 84 mathrm{~kg} \
&= 47.04 mathrm{~kg}
end{align*}
$$
So, the mass of the left side student is $47.04 mathrm{~kg}$.
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