Chapter 5
\text{Chapter 5}
Chapter 5
\text{Chapter 5}
Chapter 5
\text{Chapter 5}
Chapter 5
\text{Chapter 5}
In Figure 9, a car is driven at constant speed over a circular hill and then into a circular valley
\text{In Figure 9, a car is driven at constant speed over a circular hill and then into a circular valley}
with the same radius. At the top of the hill, the normal force on the driver from the car seat is
\text{with the same radius. At the top of the hill, the normal force on the driver from the car seat is}
zero. The driver’s mass is 80.0 kg. What is the magnitude of the normal force on the driver
\text{ zero. The driver's mass is $80.0 \mathrm{~kg}$. What is the magnitude of the normal force on the driver }
from the seat when the car passes through the bottom of the valley?
\text{from the seat when the car passes through the bottom of the valley?}
A) 1.57×103 NB) 1.01×103 NC) 1.17×103 ND) 2.81×103 NE) 2.22×103 N
\textcolor{black}{\text{A) $1.57 \times 10^3 \mathrm{~N}$}}\\
\text{B) $1.01 \times 10^3 \mathrm{~N}$}\\
\text{C) $1.17 \times 10^3 \mathrm{~N}$}\\
\text{D) $2.81 \times 10^3 \mathrm{~N}$}\\
\text{E) $2.22 \times 10^3 \mathrm{~N}$}
In Figure 9, a car is driven at constant speed over a circular hill and then into a circular valley
\text{In Figure 9, a car is driven at constant speed over a circular hill and then into a circular valley}
with the same radius. At the top of the hill, the normal force on the driver from the car seat is
\text{with the same radius. At the top of the hill, the normal force on the driver from the car seat is}
zero. The driver’s mass is 80.0 kg. What is the magnitude of the normal force on the driver
\text{ zero. The driver's mass is $80.0 \mathrm{~kg}$. What is the magnitude of the normal force on the driver }
from the seat when the car passes through the bottom of the valley?
\text{from the seat when the car passes through the bottom of the valley?}
A) 1.57×103 NB) 1.01×103 NC) 1.17×103 ND) 2.81×103 NE) 2.22×103 N
\textcolor{red}{\text{A) $1.57 \times 10^3 \mathrm{~N}$}}\\
\text{B) $1.01 \times 10^3 \mathrm{~N}$}\\
\text{C) $1.17 \times 10^3 \mathrm{~N}$}\\
\text{D) $2.81 \times 10^3 \mathrm{~N}$}\\
\text{E) $2.22 \times 10^3 \mathrm{~N}$}
∑F=ma
\sum \vec{F}=m\vec{a}
mg+FN=ma
m\vec{g}+\vec{F}_N=m\vec{a}
Let us project the eq. on the y-axis on the hill
\text{Let us project the eq. on the y-axis on the hill}
−mg+FNhill=−man
-m{g}+{F}_N^\text{hill}=-m{a}_n
FNhill=−mRv2+mg
\displaystyle {F}_N^\text{hill}=-m\frac{v^2}{R}+mg
Let us project the eq. on the y-axis on the valley
\text{Let us project the eq. on the y-axis on the valley}
−mg+FNvalley=man
-m{g}+{F}_N^\text{valley}=m{a}_n
FNvalley=mRv2+mg
\displaystyle {F}_N^\text{valley}=m\frac{v^2}{R}+mg
we sum the two equations to get:
\text{we sum the two equations to get:}
FNvalley+FNhill=2mg
\displaystyle {F}_N^\text{valley}+ {F}_N^\text{hill}=2mg
FNvalley=2mg=2×80×9.8=1.57×103N
\displaystyle {F}_N^\text{valley}=2mg=2\times 80\times9.8=1.57 \times 10^3 N
since FNhill=0,
\text{since }F_N^\text{hill}=0,
an
\vec{a}_n
an
\vec{a}_n
In Figure 6, a force F of magnitude 12 N is applied to a box of mass m2=1.0 kg. The force is
\text{In Figure 6, a force $\overrightarrow{\mathrm{F}}$ of magnitude $12 \mathrm{~N}$ is applied to a box of mass $m_2=1.0 \mathrm{~kg}$. The force is }
directed up a plane tilted by θ=37∘. The box is connected by a cord to a second box of mass
\text{directed up a plane tilted by $\theta=37^{\circ}$. The box is connected by a cord to a second box of mass}
m1=3.0 kg on the floor. The floor, plane, and pulley are frictionless, and the masses of the
\text{$m_1=3.0 \mathrm{~kg}$ on the floor. The floor, plane, and pulley are frictionless, and the masses of the }
pulley and cord are negligible. What is the tension in the cord?
\text{pulley and cord are negligible. What is the tension in the cord?}
In Figure 6, a force F of magnitude 12 N is applied to a box of mass m2=1.0 kg. The force is
\text{In Figure 6, a force $\overrightarrow{\mathrm{F}}$ of magnitude $12 \mathrm{~N}$ is applied to a box of mass $m_2=1.0 \mathrm{~kg}$. The force is }
directed up a plane tilted by θ=37∘. The box is connected by a cord to a second box of mass
\text{directed up a plane tilted by $\theta=37^{\circ}$. The box is connected by a cord to a second box of mass}
m1=3.0 kg on the floor. The floor, plane, and pulley are frictionless, and the masses of the
\text{$m_1=3.0 \mathrm{~kg}$ on the floor. The floor, plane, and pulley are frictionless, and the masses of the }
pulley and cord are negligible. What is the tension in the cord?
\text{pulley and cord are negligible. What is the tension in the cord?}
Mass m1:
\text{Mass }m_1:
T1+m1g+FN=m1a1
\vec{T}_1+m_1\vec{g}+\vec{F}_N=m_1 \vec{a}_1
Project on x-axis:
\text{Project on x-axis:}
T1=m1a1
\displaystyle {T}_1=m_1 {a_1}
Mass m2:
\text{Mass }m_2:
T2+m2g+FN+F=m2a2
\vec{T}_2+m_2\vec{g}+\vec{F}_N+\vec{F}=m_2 \vec{a}_2
Project on x-axis (incline):
\text{Project on x-axis (incline):}
−T2−m2gcos(900−θ)+F=m2a2
\displaystyle -{T}_2-m_2g \cos (90^0-\theta)+F=m_2 {a}_2
same massless corde⇒T1=T2
\text{same massless corde}\Rightarrow T_1=T_2
the two masses move together⇒a1=a2
\text{the two masses move together}\Rightarrow a_1=a_2
A◯
\text{\textcircled{A}}
B◯
\text{\textcircled{B}}
By substructing the two eq. m2A◯-m1B◯, we get:
\text{By substructing the two eq. $m_2$\textcircled{A}-$m_1$\textcircled{B}, we get:}
m2T+m1T+m2m1gcos530−m1F=0
m_2T+m_1T+m_2m_1 g \cos 53^0-m_1F=0
T=m2+m1m1(F−m2gcos530)
\displaystyle T=\frac{m_1 (F-m_2g \cos 53^0)}{m_2+m_1}
T=4.6N
\displaystyle T=4.6 N
T1
\vec{T}_1
T2
\vec{T}_2
m1g
m_1 \vec{g}
m2g
m_2 \vec{g}
FN
\vec{F}_N
FN
\vec{F}_N
Two masses m1=2.0 kg and m2=3.0 kg are connected as shown in Fig 4.
\text{Two masses $\mathrm{m}_1=2.0 \mathrm{~kg}$ and $\mathrm{m}_2=3.0 \mathrm{~kg}$ are connected as shown in Fig 4.}
Find the tension T2 if the tension T1= 50.0 N.
\text{ Find the tension $\mathrm{T}_2$ if the tension $\mathrm{T}_1=$ $50.0 \mathrm{~N}$.}
A) zeroB) 50.0 NC) 20.0 ND) 30.0 NE) 10.0 N
\text{A) zero}\\
\text{B) $50.0 \mathrm{~N}$}\\
\text{C) $20.0 \mathrm{~N}$}\\
\text{D) $30.0 \mathrm{~N}$}\\
\text{E) $10.0 \mathrm{~N}$}
Two masses m1=2.0 kg and m2=3.0 kg are connected as shown in Fig 4.
\text{Two masses $\mathrm{m}_1=2.0 \mathrm{~kg}$ and $\mathrm{m}_2=3.0 \mathrm{~kg}$ are connected as shown in Fig 4.}
Find the tension T2 if the tension T1= 50.0 N.
\text{ Find the tension $\mathrm{T}_2$ if the tension $\mathrm{T}_1=$ $50.0 \mathrm{~N}$.}
On m2
\text{On $m_2$}
T2+m2g+N=m2a
\vec{T}_2+m_2\vec{g}+\vec{N}=m_2\vec{a}
We project on x-axis
\text{We project on x-axis}
T2=m2a
{T}_2=m_2{a}
On m1
\text{On $m_1$}
T1=(m1+m2)a
\vec{T}_1=(m_1+m_2)\vec{a}
We project on x-axis
\text{We project on x-axis}
T1=(m1+m2)a
{T}_1=(m_1+m_2){a}
because T1 is pulling both masses (no need to plot N and mg)
\text{because $T_1$ is pulling both masses (no need to plot $\vec{N}$ and m$\vec{g}$)}
⇒
\Rightarrow
a=(m1+m2)T1
\displaystyle a=\frac{{T}_1}{(m_1+m_2)}
⇒
\Rightarrow
T2=(m1+m2)m2T1=30N
\displaystyle T_2=\frac{m_2{T}_1}{(m_1+m_2)}=30N
A) zeroB) 50.0 NC) 20.0 ND) 30.0 NE) 10.0 N
\text{A) zero}\\
\text{B) $50.0 \mathrm{~N}$}\\
\text{C) $20.0 \mathrm{~N}$}\\
\text{\textcolor{red}{D) $30.0 \mathrm{~N}$}}\\
\text{E) $10.0 \mathrm{~N}$}
Two masses m1=2.0 kg and m2=3.0 kg are connected as shown in Fig 4.
\text{Two masses $\mathrm{m}_1=2.0 \mathrm{~kg}$ and $\mathrm{m}_2=3.0 \mathrm{~kg}$ are connected as shown in Fig 4.}
Find the tension T2 if the tension T1= 50.0 N.
\text{ Find the tension $\mathrm{T}_2$ if the tension $\mathrm{T}_1=$ $50.0 \mathrm{~N}$.}
On m2
\text{On $m_2$}
T2+m2g+N=m2a
\vec{T}_2+m_2\vec{g}+\vec{N}=m_2\vec{a}
We project on x-axis
\text{We project on x-axis}
T2=m2a
{T}_2=m_2{a}
On m1
\text{On $m_1$}
T1=(m1+m2)a
\vec{T}_1=(m_1+m_2)\vec{a}
A) zeroB) 50.0 NC) 20.0 ND) 30.0 NE) 10.0 N
\text{A) zero}\\
\text{B) $50.0 \mathrm{~N}$}\\
\text{C) $20.0 \mathrm{~N}$}\\
\text{\textcolor{red}{D) $30.0 \mathrm{~N}$}}\\
\text{E) $10.0 \mathrm{~N}$}
Instead of the above equation, we could have wrote:
\text{Instead of the above equation, we could have wrote:}
T1+T2=m1a
\vec{T}_1+\vec{T}_2=m_1\vec{a}
⇒
\Rightarrow
T1−T2=m1a
{T}_1-{T}_2=m_1{a}
and then we use the first equation with T1
\text{and then we use the first equation with }T_1
Another way of solving
\text{Another way of solving}
Two masses m1(=2.0 kg) and m2(=3.0 kg) are connected as shown in Fig 4.
\text{Two masses $\mathrm{m} 1(=2.0 \mathrm{~kg})$ and $\mathrm{m} 2(=3.0 \mathrm{~kg})$ are connected as shown in Fig 4.}
Find the tension T2 if the tension T1= 50.0 N.
\text{ Find the tension $\mathrm{T} 2$ if the tension $\mathrm{T} 1=$ $50.0 \mathrm{~N}$.}
360
36^0
α
\alpha
C h a p t e r 5 \text{Chapter 5} C h a p t e r 5 \text{Chapter 5} C h a p t e r 5 \text{Chapter 5} C h a p t e r 5 \text{Chapter 5}
chapter 5-Phys101
By smstry
chapter 5-Phys101
- 163
