Quiz 4

Physics 6A

2/5/03

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 shown below are connected by a massless string and massless pulley. The block on the inclined plane has a mass $m_1 = 7 kg$ and the block that is hanging has mass $m_2 = 8 kg$. The inclined plane makes an angle $\theta = 20$ degrees with the horizontal. The coefficient of kinetic friction is $\mu_k = 0.15$. Take the acceleration of gravity to be $9.8 m/s^2$ and assume that the velocity and the acceleration of $m_1$ are up the inclined plane in the direction shown by vector $a$.

(a) (5 points) Draw two free body diagrams. One for $m_1$, and one for $m_2$.

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

(c) (5 points) What is the tension in the string?

Solution

(a): Free body diagrams:

(b) get the acceleration

F=ma in component form

We've chosen the coordinate systems as shown.

Let's look at the first mass. We'll denote its acceleration by ${\bf a}_1$.

We have ${\bf F}_{net} ~=~ m{\bf a}_1$ or ${\bf N}+ {\bf W}_1 + {\bf T}+{\bf f} ~=~ m{\bf a}$

The decomposition into components for the weight $W_1$ is the same as in the web book. So there are two forces acting in the y direction. The weight has a component $W_{1y}$ in this direction and the normal force entirely in this direction. The acceleration in the y direction is zero, since the block is sliding along the x direction. So the y component of the above equation is

$$N ~=~ W_1 \cos \theta$$ (1)

Here $W_1 = m_1 g = 68.60$.

Numerically, $N = 64.46 N$.

The x component is $T - f - W_1 \sin \theta = m_1 a_1$

But we also know that $f ~=~ \mu_k N = 9.67 N$. Putting this all together, the y equation becomes

$$T-\mu_k W_1 \cos\theta - W_1 \sin \theta = m_1a_1$$ (2)

Now on to block 2. Here all the action occurs in the y direction, We have

$$T-W_2 ~=~ m_2 a_2$$ (3)

Here $W_2 = m_2 g = 78.40$.

If the first mass goes down the plane one inch, the second mass must go up the same amount. So the accelerations are equal and opposite (think about the coordinate systems we're using)

$$a_1 ~=~ -a_2$$ (4)

Solve the equations

Now we have a third equation so can go ahead and solve for the acceleration. Subtracting the above equations we can eliminate the tension giving

$$W_2 -\mu_k W_1 \cos\theta - W_1 \sin \theta ~=~ m_1 a_1 - m_2 a_2 ~=~ (m_1+m_2)a_1$$ (5)

Now remembering that $W_1 = m_1 g$ and $W_2 = m_2 g$, we can solve for the acceleration

$$a = a_1 ~=~ {g\over (m_1+m_2)} \large [ m_2 -(\mu_k \cos\theta +\sin\theta)m_1)\large ]$$ (6)

But $\cos(\theta) = 0.94$ and $\sin(\theta) = 0.34$.

So numerically this is $a = 3.02m/s^2$.

(c) get the tension

Now we can use the acceleration to obtain the tension since from above $T-W_2 ~=~ m_2 a_2 = -m_2 a$. Therefore $T = m_2 g - m_2 a = m_2(g-a) = 54.26N$.

Quiz 4

Physics 6A

2/5/03

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 shown below are connected by a massless string and massless pulley. The block on the inclined plane has a mass $m_1 = 6 kg$ and the block that is hanging has mass $m_2 = 7 kg$. The inclined plane makes an angle $\theta = 34$ degrees with the horizontal. The coefficient of kinetic friction is $\mu_k = 0.1$. Take the acceleration of gravity to be $9.8 m/s^2$ and assume that the velocity and the acceleration of $m_1$ are up the inclined plane in the direction shown by vector $a$.

(a) (5 points) Draw two free body diagrams. One for $m_1$, and one for $m_2$.

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

(c) (5 points) What is the tension in the string?

Solution

(a): Free body diagrams:

(b) get the acceleration

F=ma in component form

We've chosen the coordinate systems as shown.

Let's look at the first mass. We'll denote its acceleration by ${\bf a}_1$.

We have ${\bf F}_{net} ~=~ m{\bf a}_1$ or ${\bf N}+ {\bf W}_1 + {\bf T}+{\bf f} ~=~ m{\bf a}$

The decomposition into components for the weight $W_1$ is the same as in the web book. So there are two forces acting in the y direction. The weight has a component $W_{1y}$ in this direction and the normal force entirely in this direction. The acceleration in the y direction is zero, since the block is sliding along the x direction. So the y component of the above equation is

$$N ~=~ W_1 \cos \theta$$ (7)

Here $W_1 = m_1 g = 58.80$.

Numerically, $N = 48.75 N$.

The x component is $T - f - W_1 \sin \theta = m_1 a_1$

But we also know that $f ~=~ \mu_k N = 4.87 N$. Putting this all together, the y equation becomes

$$T-\mu_k W_1 \cos\theta - W_1 \sin \theta = m_1a_1$$ (8)

Now on to block 2. Here all the action occurs in the y direction, We have

$$T-W_2 ~=~ m_2 a_2$$ (9)

Here $W_2 = m_2 g = 68.60$.

If the first mass goes down the plane one inch, the second mass must go up the same amount. So the accelerations are equal and opposite (think about the coordinate systems we're using)

$$a_1 ~=~ -a_2$$ (10)

Solve the equations

Now we have a third equation so can go ahead and solve for the acceleration. Subtracting the above equations we can eliminate the tension giving

$$W_2 -\mu_k W_1 \cos\theta - W_1 \sin \theta ~=~ m_1 a_1 - m_2 a_2 ~=~ (m_1+m_2)a_1$$ (11)

Now remembering that $W_1 = m_1 g$ and $W_2 = m_2 g$, we can solve for the acceleration

$$a = a_1 ~=~ {g\over (m_1+m_2)} \large [ m_2 -(\mu_k \cos\theta +\sin\theta)m_1)\large ]$$ (12)

But $\cos(\theta) = 0.83$ and $\sin(\theta) = 0.56$.

So numerically this is $a = 2.37m/s^2$.

(c) get the tension

Now we can use the acceleration to obtain the tension since from above $T-W_2 ~=~ m_2 a_2 = -m_2 a$. Therefore $T = m_2 g - m_2 a = m_2(g-a) = 51.99N$.

Quiz 4

Physics 6A

2/5/03

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 shown below are connected by a massless string and massless pulley. The block on the inclined plane has a mass $m_1 = 5 kg$ and the block that is hanging has mass $m_2 = 8 kg$. The inclined plane makes an angle $\theta = 29$ degrees with the horizontal. The coefficient of kinetic friction is $\mu_k = 0.25$. Take the acceleration of gravity to be $9.8 m/s^2$ and assume that the velocity and the acceleration of $m_1$ are up the inclined plane in the direction shown by vector $a$.

(a) (5 points) Draw two free body diagrams. One for $m_1$, and one for $m_2$.

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

(c) (5 points) What is the tension in the string?

Solution

(a): Free body diagrams:

(b) get the acceleration

F=ma in component form

We've chosen the coordinate systems as shown.

Let's look at the first mass. We'll denote its acceleration by ${\bf a}_1$.

We have ${\bf F}_{net} ~=~ m{\bf a}_1$ or ${\bf N}+ {\bf W}_1 + {\bf T}+{\bf f} ~=~ m{\bf a}$

The decomposition into components for the weight $W_1$ is the same as in the web book. So there are two forces acting in the y direction. The weight has a component $W_{1y}$ in this direction and the normal force entirely in this direction. The acceleration in the y direction is zero, since the block is sliding along the x direction. So the y component of the above equation is

$$N ~=~ W_1 \cos \theta$$ (13)

Here $W_1 = m_1 g = 49.00$.

Numerically, $N = 42.86 N$.

The x component is $T - f - W_1 \sin \theta = m_1 a_1$

But we also know that $f ~=~ \mu_k N = 10.71 N$. Putting this all together, the y equation becomes

$$T-\mu_k W_1 \cos\theta - W_1 \sin \theta = m_1a_1$$ (14)

Now on to block 2. Here all the action occurs in the y direction, We have

$$T-W_2 ~=~ m_2 a_2$$ (15)

Here $W_2 = m_2 g = 78.40$.

If the first mass goes down the plane one inch, the second mass must go up the same amount. So the accelerations are equal and opposite (think about the coordinate systems we're using)

$$a_1 ~=~ -a_2$$ (16)

Solve the equations

Now we have a third equation so can go ahead and solve for the acceleration. Subtracting the above equations we can eliminate the tension giving

$$W_2 -\mu_k W_1 \cos\theta - W_1 \sin \theta ~=~ m_1 a_1 - m_2 a_2 ~=~ (m_1+m_2)a_1$$ (17)

Now remembering that $W_1 = m_1 g$ and $W_2 = m_2 g$, we can solve for the acceleration

$$a = a_1 ~=~ {g\over (m_1+m_2)} \large [ m_2 -(\mu_k \cos\theta +\sin\theta)m_1)\large ]$$ (18)

But $\cos(\theta) = 0.87$ and $\sin(\theta) = 0.48$.

So numerically this is $a = 3.38m/s^2$.

(c) get the tension

Now we can use the acceleration to obtain the tension since from above $T-W_2 ~=~ m_2 a_2 = -m_2 a$. Therefore $T = m_2 g - m_2 a = m_2(g-a) = 51.37N$.

Quiz 4

Physics 6A

2/5/03

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 shown below are connected by a massless string and massless pulley. The block on the inclined plane has a mass $m_1 = 5 kg$ and the block that is hanging has mass $m_2 = 6 kg$. The inclined plane makes an angle $\theta = 23$ degrees with the horizontal. The coefficient of kinetic friction is $\mu_k = 0.19$. Take the acceleration of gravity to be $9.8 m/s^2$ and assume that the velocity and the acceleration of $m_1$ are up the inclined plane in the direction shown by vector $a$.

(a) (5 points) Draw two free body diagrams. One for $m_1$, and one for $m_2$.

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

(c) (5 points) What is the tension in the string?

Solution

(a): Free body diagrams:

(b) get the acceleration

F=ma in component form

We've chosen the coordinate systems as shown.

Let's look at the first mass. We'll denote its acceleration by ${\bf a}_1$.

We have ${\bf F}_{net} ~=~ m{\bf a}_1$ or ${\bf N}+ {\bf W}_1 + {\bf T}+{\bf f} ~=~ m{\bf a}$

The decomposition into components for the weight $W_1$ is the same as in the web book. So there are two forces acting in the y direction. The weight has a component $W_{1y}$ in this direction and the normal force entirely in this direction. The acceleration in the y direction is zero, since the block is sliding along the x direction. So the y component of the above equation is

$$N ~=~ W_1 \cos \theta$$ (19)

Here $W_1 = m_1 g = 49.00$.

Numerically, $N = 45.10 N$.

The x component is $T - f - W_1 \sin \theta = m_1 a_1$

But we also know that $f ~=~ \mu_k N = 8.57 N$. Putting this all together, the y equation becomes

$$T-\mu_k W_1 \cos\theta - W_1 \sin \theta = m_1a_1$$ (20)

Now on to block 2. Here all the action occurs in the y direction, We have

$$T-W_2 ~=~ m_2 a_2$$ (21)

Here $W_2 = m_2 g = 58.80$.

If the first mass goes down the plane one inch, the second mass must go up the same amount. So the accelerations are equal and opposite (think about the coordinate systems we're using)

$$a_1 ~=~ -a_2$$ (22)

Solve the equations

Now we have a third equation so can go ahead and solve for the acceleration. Subtracting the above equations we can eliminate the tension giving

$$W_2 -\mu_k W_1 \cos\theta - W_1 \sin \theta ~=~ m_1 a_1 - m_2 a_2 ~=~ (m_1+m_2)a_1$$ (23)

Now remembering that $W_1 = m_1 g$ and $W_2 = m_2 g$, we can solve for the acceleration

$$a = a_1 ~=~ {g\over (m_1+m_2)} \large [ m_2 -(\mu_k \cos\theta +\sin\theta)m_1)\large ]$$ (24)

But $\cos(\theta) = 0.92$ and $\sin(\theta) = 0.39$.

So numerically this is $a = 2.83m/s^2$.

(c) get the tension

Now we can use the acceleration to obtain the tension since from above $T-W_2 ~=~ m_2 a_2 = -m_2 a$. Therefore $T = m_2 g - m_2 a = m_2(g-a) = 41.84N$.

Quiz 4

Physics 6A

2/5/03

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 shown below are connected by a massless string and massless pulley. The block on the inclined plane has a mass $m_1 = 4 kg$ and the block that is hanging has mass $m_2 = 7 kg$. The inclined plane makes an angle $\theta = 37$ degrees with the horizontal. The coefficient of kinetic friction is $\mu_k = 0.14$. Take the acceleration of gravity to be $9.8 m/s^2$ and assume that the velocity and the acceleration of $m_1$ are up the inclined plane in the direction shown by vector $a$.

(a) (5 points) Draw two free body diagrams. One for $m_1$, and one for $m_2$.

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

(c) (5 points) What is the tension in the string?

Solution

(a): Free body diagrams:

(b) get the acceleration

F=ma in component form

We've chosen the coordinate systems as shown.

Let's look at the first mass. We'll denote its acceleration by ${\bf a}_1$.

We have ${\bf F}_{net} ~=~ m{\bf a}_1$ or ${\bf N}+ {\bf W}_1 + {\bf T}+{\bf f} ~=~ m{\bf a}$

The decomposition into components for the weight $W_1$ is the same as in the web book. So there are two forces acting in the y direction. The weight has a component $W_{1y}$ in this direction and the normal force entirely in this direction. The acceleration in the y direction is zero, since the block is sliding along the x direction. So the y component of the above equation is

$$N ~=~ W_1 \cos \theta$$ (25)

Here $W_1 = m_1 g = 39.20$.

Numerically, $N = 31.31 N$.

The x component is $T - f - W_1 \sin \theta = m_1 a_1$

But we also know that $f ~=~ \mu_k N = 4.38 N$. Putting this all together, the y equation becomes

$$T-\mu_k W_1 \cos\theta - W_1 \sin \theta = m_1a_1$$ (26)

Now on to block 2. Here all the action occurs in the y direction, We have

$$T-W_2 ~=~ m_2 a_2$$ (27)

Here $W_2 = m_2 g = 68.60$.

If the first mass goes down the plane one inch, the second mass must go up the same amount. So the accelerations are equal and opposite (think about the coordinate systems we're using)

$$a_1 ~=~ -a_2$$ (28)

Solve the equations

Now we have a third equation so can go ahead and solve for the acceleration. Subtracting the above equations we can eliminate the tension giving

$$W_2 -\mu_k W_1 \cos\theta - W_1 \sin \theta ~=~ m_1 a_1 - m_2 a_2 ~=~ (m_1+m_2)a_1$$ (29)

Now remembering that $W_1 = m_1 g$ and $W_2 = m_2 g$, we can solve for the acceleration

$$a = a_1 ~=~ {g\over (m_1+m_2)} \large [ m_2 -(\mu_k \cos\theta +\sin\theta)m_1)\large ]$$ (30)

But $\cos(\theta) = 0.80$ and $\sin(\theta) = 0.60$.

So numerically this is $a = 3.69m/s^2$.

(c) get the tension

Now we can use the acceleration to obtain the tension since from above $T-W_2 ~=~ m_2 a_2 = -m_2 a$. Therefore $T = m_2 g - m_2 a = m_2(g-a) = 42.75N$.

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