\text{Chapter 5}
\text{Chapter 5}
\text{Chapter 5}
\text{Chapter 5}
\text{In Figure 9, a car is driven at constant speed over a circular hill and then into a circular valley}
\text{with the same radius. At the top of the hill, the normal force on the driver from the car seat is}
\text{ zero. The driver's mass is $80.0 \mathrm{~kg}$. What is the magnitude of the normal force on the driver }
\text{from the seat when the car passes through the bottom of the valley?}
\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) $2.22 \times 10^3 \mathrm{~N}$}
\text{In Figure 9, a car is driven at constant speed over a circular hill and then into a circular valley}
\text{with the same radius. At the top of the hill, the normal force on the driver from the car seat is}
\text{ zero. The driver's mass is $80.0 \mathrm{~kg}$. What is the magnitude of the normal force on the driver }
\text{from the seat when the car passes through the bottom of the valley?}
\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) $2.22 \times 10^3 \mathrm{~N}$}
\sum \vec{F}=m\vec{a}
m\vec{g}+\vec{F}_N=m\vec{a}
\text{Let us project the eq. on the y-axis on the hill}
-m{g}+{F}_N^\text{hill}=-m{a}_n
\displaystyle {F}_N^\text{hill}=-m\frac{v^2}{R}+mg
\text{Let us project the eq. on the y-axis on the valley}
-m{g}+{F}_N^\text{valley}=m{a}_n
\displaystyle {F}_N^\text{valley}=m\frac{v^2}{R}+mg
\text{we sum the two equations to get:}
\displaystyle {F}_N^\text{valley}+ {F}_N^\text{hill}=2mg
\displaystyle {F}_N^\text{valley}=2mg=2\times 80\times9.8=1.57 \times 10^3 N
\text{since }F_N^\text{hill}=0,
\vec{a}_n
\vec{a}_n
\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 }
\text{directed up a plane tilted by $\theta=37^{\circ}$. The box is connected by a cord to a second box of mass}
\text{$m_1=3.0 \mathrm{~kg}$ on the floor. The floor, plane, and pulley are frictionless, and the masses of the }
\text{pulley and cord are negligible. What is the tension in the cord?}
\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 }
\text{directed up a plane tilted by $\theta=37^{\circ}$. The box is connected by a cord to a second box of mass}
\text{$m_1=3.0 \mathrm{~kg}$ on the floor. The floor, plane, and pulley are frictionless, and the masses of the }
\text{pulley and cord are negligible. What is the tension in the cord?}
\text{Mass }m_1:
\vec{T}_1+m_1\vec{g}+\vec{F}_N=m_1 \vec{a}_1
\text{Project on x-axis:}
\displaystyle {T}_1=m_1 {a_1}
\text{Mass }m_2:
\vec{T}_2+m_2\vec{g}+\vec{F}_N+\vec{F}=m_2 \vec{a}_2
\text{Project on x-axis (incline):}
\displaystyle -{T}_2-m_2g \cos (90^0-\theta)+F=m_2 {a}_2
\text{same massless corde}\Rightarrow T_1=T_2
\text{the two masses move together}\Rightarrow a_1=a_2
\text{\textcircled{A}}
\text{\textcircled{B}}
\text{By substructing the two eq. $m_2$\textcircled{A}-$m_1$\textcircled{B}, we get:}
m_2T+m_1T+m_2m_1 g \cos 53^0-m_1F=0
\displaystyle T=\frac{m_1 (F-m_2g \cos 53^0)}{m_2+m_1}
\displaystyle T=4.6 N
\vec{T}_1
\vec{T}_2
m_1 \vec{g}
m_2 \vec{g}
\vec{F}_N
\vec{F}_N
\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.}
\text{ Find the tension $\mathrm{T}_2$ if the tension $\mathrm{T}_1=$ $50.0 \mathrm{~N}$.}
\text{A) zero}\\
\text{B) $50.0 \mathrm{~N}$}\\
\text{C) $20.0 \mathrm{~N}$}\\
\text{D) $30.0 \mathrm{~N}$}\\
\text{E) $10.0 \mathrm{~N}$}
\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.}
\text{ Find the tension $\mathrm{T}_2$ if the tension $\mathrm{T}_1=$ $50.0 \mathrm{~N}$.}
\text{On $m_2$}
\vec{T}_2+m_2\vec{g}+\vec{N}=m_2\vec{a}
\text{We project on x-axis}
{T}_2=m_2{a}
\text{On $m_1$}
\vec{T}_1=(m_1+m_2)\vec{a}
\text{We project on x-axis}
{T}_1=(m_1+m_2){a}
\text{because $T_1$ is pulling both masses (no need to plot $\vec{N}$ and m$\vec{g}$)}
\Rightarrow
\displaystyle a=\frac{{T}_1}{(m_1+m_2)}
\Rightarrow
\displaystyle T_2=\frac{m_2{T}_1}{(m_1+m_2)}=30N
\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) $10.0 \mathrm{~N}$}
\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.}
\text{ Find the tension $\mathrm{T}_2$ if the tension $\mathrm{T}_1=$ $50.0 \mathrm{~N}$.}
\text{On $m_2$}
\vec{T}_2+m_2\vec{g}+\vec{N}=m_2\vec{a}
\text{We project on x-axis}
{T}_2=m_2{a}
\text{On $m_1$}
\vec{T}_1=(m_1+m_2)\vec{a}
\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) $10.0 \mathrm{~N}$}
\text{Instead of the above equation, we could have wrote:}
\vec{T}_1+\vec{T}_2=m_1\vec{a}
\Rightarrow
{T}_1-{T}_2=m_1{a}
\text{and then we use the first equation with }T_1
\text{Another way of solving}
\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.}
\text{ Find the tension $\mathrm{T} 2$ if the tension $\mathrm{T} 1=$ $50.0 \mathrm{~N}$.}
36^0
\alpha