Quiz 3

Physics 6A

1/29/03

Please read the question carefully before attempting it. You will not be given any credit if you only write down the final answer. You must show your work.

The two blocks below are connected by a heavy rope of mass $m_r = 5 kg$. The top block has a mass $m_t = 4 kg$ and the bottom block has mass $m_b = 6 kg$. An upward force of $F = 188 N$ is applied as shown. Take the acceleration of gravity to be $9.8 m/s^2$.

(a) (5 points) Draw three free body diagrams. One for the top block, one for the rope and one for the bottom block. For each force, indicate which body exerts that force.

(b) (5 points) What is the acceleration of the system?

(c) (5 points) What is the tension at the top of the heavy rope?

Solution

(a)

(b)

All objects move together and so all have the same acceleration. On top block $F_{net} = m_t a$. but $F_{net} = F - T_t - m_t g$

On bottom block $F_{net} = m_b a$. here with $F_{net} = T_b - m_b g$

On rope block $F_{net} = m_r a$. with $F_{net} = T_t - T_b - m_r g$

Add them all together: $(m_t + m_b + m_r)a = (F-T_t-m_t g) + (T_b- m_b g) + (T_t - T_b -m_r g)$.

Cancelling we have

$(m_t + m_b + m_r)a = (F-m_t g) + (- m_b g) + ( -m_r g)$. So $a = (F/m_{total} -g)$ where $m_{total} = m_t + m_b + m_r = 15$. So $a = 2.73 m/s^2$.

(c)

Solving for $T_t$, $T_t = F - F_{net} - m_t g = F - m_t a -m_t g = F - m_t(a+g) = F - m_t(F/m_{tota... ..._t/m_{total}) = F((m_t+m_r+m_b) -m_t)/m_{total} =F(m_r+m_b)/m_{total} = 137.87N$.

Quiz 3

Physics 6A

1/29/03

Please read the question carefully before attempting it. You will not be given any credit if you only write down the final answer. You must show your work.

The two blocks below are connected by a heavy rope of mass $m_r = 4 kg$. The top block has a mass $m_t = 3 kg$ and the bottom block has mass $m_b = 9 kg$. An upward force of $F = 181.8 N$ is applied as shown. Take the acceleration of gravity to be $9.8 m/s^2$.

(a) (5 points) Draw three free body diagrams. One for the top block, one for the rope and one for the bottom block. For each force, indicate which body exerts that force.

(b) (5 points) What is the acceleration of the system?

(c) (5 points) What is the tension at the top of the heavy rope?

Solution

(a)

(b)

All objects move together and so all have the same acceleration. On top block $F_{net} = m_t a$. but $F_{net} = F - T_t - m_t g$

On bottom block $F_{net} = m_b a$. here with $F_{net} = T_b - m_b g$

On rope block $F_{net} = m_r a$. with $F_{net} = T_t - T_b - m_r g$

Add them all together: $(m_t + m_b + m_r)a = (F-T_t-m_t g) + (T_b- m_b g) + (T_t - T_b -m_r g)$.

Cancelling we have

$(m_t + m_b + m_r)a = (F-m_t g) + (- m_b g) + ( -m_r g)$. So $a = (F/m_{total} -g)$ where $m_{total} = m_t + m_b + m_r = 16$. So $a = 1.56 m/s^2$.

(c)

Solving for $T_t$, $T_t = F - F_{net} - m_t g = F - m_t a -m_t g = F - m_t(a+g) = F - m_t(F/m_{tota... ..._t/m_{total}) = F((m_t+m_r+m_b) -m_t)/m_{total} =F(m_r+m_b)/m_{total} = 147.71N$.

Quiz 3

Physics 6A

1/29/03

Please read the question carefully before attempting it. You will not be given any credit if you only write down the final answer. You must show your work.

The two blocks below are connected by a heavy rope of mass $m_r = 2 kg$. The top block has a mass $m_t = 7 kg$ and the bottom block has mass $m_b = 8 kg$. An upward force of $F = 205.6 N$ is applied as shown. Take the acceleration of gravity to be $9.8 m/s^2$.

(a) (5 points) Draw three free body diagrams. One for the top block, one for the rope and one for the bottom block. For each force, indicate which body exerts that force.

(b) (5 points) What is the acceleration of the system?

(c) (5 points) What is the tension at the top of the heavy rope?

Solution

(a)

(b)

All objects move together and so all have the same acceleration. On top block $F_{net} = m_t a$. but $F_{net} = F - T_t - m_t g$

On bottom block $F_{net} = m_b a$. here with $F_{net} = T_b - m_b g$

On rope block $F_{net} = m_r a$. with $F_{net} = T_t - T_b - m_r g$

Add them all together: $(m_t + m_b + m_r)a = (F-T_t-m_t g) + (T_b- m_b g) + (T_t - T_b -m_r g)$.

Cancelling we have

$(m_t + m_b + m_r)a = (F-m_t g) + (- m_b g) + ( -m_r g)$. So $a = (F/m_{total} -g)$ where $m_{total} = m_t + m_b + m_r = 17$. So $a = 2.29 m/s^2$.

(c)

Solving for $T_t$, $T_t = F - F_{net} - m_t g = F - m_t a -m_t g = F - m_t(a+g) = F - m_t(F/m_{tota... ..._t/m_{total}) = F((m_t+m_r+m_b) -m_t)/m_{total} =F(m_r+m_b)/m_{total} = 120.94N$.

Quiz 3

Physics 6A

1/29/03

Please read the question carefully before attempting it. You will not be given any credit if you only write down the final answer. You must show your work.

The two blocks below are connected by a heavy rope of mass $m_r = 1 kg$. The top block has a mass $m_t = 7 kg$ and the bottom block has mass $m_b = 7 kg$. An upward force of $F = 170 N$ is applied as shown. Take the acceleration of gravity to be $9.8 m/s^2$.

(a) (5 points) Draw three free body diagrams. One for the top block, one for the rope and one for the bottom block. For each force, indicate which body exerts that force.

(b) (5 points) What is the acceleration of the system?

(c) (5 points) What is the tension at the top of the heavy rope?

Solution

(a)

(b)

All objects move together and so all have the same acceleration. On top block $F_{net} = m_t a$. but $F_{net} = F - T_t - m_t g$

On bottom block $F_{net} = m_b a$. here with $F_{net} = T_b - m_b g$

On rope block $F_{net} = m_r a$. with $F_{net} = T_t - T_b - m_r g$

Add them all together: $(m_t + m_b + m_r)a = (F-T_t-m_t g) + (T_b- m_b g) + (T_t - T_b -m_r g)$.

Cancelling we have

$(m_t + m_b + m_r)a = (F-m_t g) + (- m_b g) + ( -m_r g)$. So $a = (F/m_{total} -g)$ where $m_{total} = m_t + m_b + m_r = 15$. So $a = 1.53 m/s^2$.

(c)

Solving for $T_t$, $T_t = F - F_{net} - m_t g = F - m_t a -m_t g = F - m_t(a+g) = F - m_t(F/m_{tota... ...m_t/m_{total}) = F((m_t+m_r+m_b) -m_t)/m_{total} =F(m_r+m_b)/m_{total} = 90.67N$.

Quiz 3

Physics 6A

1/29/03

Please read the question carefully before attempting it. You will not be given any credit if you only write down the final answer. You must show your work.

The two blocks below are connected by a heavy rope of mass $m_r = 4 kg$. The top block has a mass $m_t = 6 kg$ and the bottom block has mass $m_b = 5 kg$. An upward force of $F = 184 N$ is applied as shown. Take the acceleration of gravity to be $9.8 m/s^2$.

(a) (5 points) Draw three free body diagrams. One for the top block, one for the rope and one for the bottom block. For each force, indicate which body exerts that force.

(b) (5 points) What is the acceleration of the system?

(c) (5 points) What is the tension at the top of the heavy rope?

Solution

(a)

(b)

All objects move together and so all have the same acceleration. On top block $F_{net} = m_t a$. but $F_{net} = F - T_t - m_t g$

On bottom block $F_{net} = m_b a$. here with $F_{net} = T_b - m_b g$

On rope block $F_{net} = m_r a$. with $F_{net} = T_t - T_b - m_r g$

Add them all together: $(m_t + m_b + m_r)a = (F-T_t-m_t g) + (T_b- m_b g) + (T_t - T_b -m_r g)$.

Cancelling we have

$(m_t + m_b + m_r)a = (F-m_t g) + (- m_b g) + ( -m_r g)$. So $a = (F/m_{total} -g)$ where $m_{total} = m_t + m_b + m_r = 15$. So $a = 2.47 m/s^2$.

(c)

Solving for $T_t$, $T_t = F - F_{net} - m_t g = F - m_t a -m_t g = F - m_t(a+g) = F - m_t(F/m_{tota... ..._t/m_{total}) = F((m_t+m_r+m_b) -m_t)/m_{total} =F(m_r+m_b)/m_{total} = 110.40N$.

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