- #1
RoyalCat
- 671
- 2
Case closed. :)
Well, I've got quite a few questions I couldn't solve. I'd love to be pointed in the right direction, because there are some problems I just don't know how to approach at all.
This is just the first batch, I'll be adding a couple more in a while.
I've attached a link to the relevant diagrams at the bottom of the post. Images removed pending approval.
Question 1
A fly is standing on the bottom of a close jar, positioned on one side of a sensitive balance scale which is initially leveled.
Suddenly, the fly rises up, hovers in place for several seconds, and touches back down.
Which of the following sentences most accurately describes what happens?
A. The scales remain leveled all the time.
B. The side with the jar will go down once the fly rises, stay leveled while the fly hovers in place, and come back up when the fly comes down.
C. The side with the jar will go up once the fly rises, stay leveled while the fly hovers in place, and come back when the fly comes down.
D. The side with the jar will go down all the while the fly is in the air.
The answer according to the book is B.
I am completely lost. Why would it go down if the fly is applying LESS force..? And why would it stay leveled if it's still imbalanced? Bleh, I can't make heads or tails of it.
Question 2
In the system in the diagram, the two masses, corresponding to [tex]W_1[/tex] and [tex]W_2[/tex] are given.
The coefficient of kinetic friction between the floor and the cart is [tex]\mu[/tex]
There is no friction between the two masses.
It is given that the cart is traveling with a constant velocity, [tex]V_1[/tex] to the right, and that the weight is descending at a constant velocity [tex]V_2[/tex]
A. Calculate the tension in the string.
B. Calculate the coefficient of friction, [tex]\mu[/tex]
C. What is the magnitude of the horizontal force which the cart exerts on the weight?
D. Calculate the ratio between the horizontal and vertical components of the weight's velocity, [tex]\frac{V_{2x}}{V_{2y}}[/tex]
E. How would the above results change if the coefficient of friction between the cart and the weight weren't 0?
My attempt:
The forces acting on the weight are:
Its weight, [tex]W_2[/tex]
The tension from the string, [tex]T_[/tex]
The normal force from the cart, [tex]N_{cw}[/tex]
Since it's moving at a constant velocity, then [tex]\Sigma \vec F=0[/tex]
Therefore:
[tex]N_{cw}=0[/tex] (This is the answer to C, and the book agrees with me on it.)
[tex]T=W_2[/tex] (This is the answer to A, and the book agrees with me on it)
The forces acting on the cart are:
Its weight, [tex]W_1[/tex]
The normal force from the floor, [tex]N[/tex]
The tension in the string, acting on the right of the cart, where its connected to it, [tex]T[/tex]
The kinetic friction, [tex]\mu N[/tex]
The tension in the string again, acting on the pulley on the top of the cart, [tex]T[/tex]
The normal force from the weight, [tex]N_{wc}=0[/tex]
Since it's moving at a constant velocity, then [tex]\Sigma \vec F=0[/tex]
Therefore:
[tex]N=W_1+T=W_1+W_2[/tex]
[tex]f_k=\mu(W_1+W_2)[/tex]
[tex]T=f_k[/tex]
Equating the two expressions for the the tension, I receive:
[tex]\mu=\frac{W_2}{W_1+W_2}[/tex]
The book's answer is:
[tex]\mu=\frac{2W_2}{W_1+W_2}[/tex]
Where did I go wrong so far? Should I have counted the tension a third time as well?
For every meter the weight descends, it pulls the string, and the cart, one meter to the right. So the ratio [tex]\frac{V_{2x}}{V_{2y}}[/tex] is 1.
The book's answer, however, is 1:2. I can't understand this.
Since the normal force between the weight and the cart is 0, any resulting friction would be 0, therefore, not changing the result by one bit.
Question 3
A uniform rod is placed horizontally at equilibrium on a smooth axis through its center. Two masses, [tex]m[/tex] are suspended on the rod as illustrated.
A. Will the rod remain horizontal? If not, which way will it turn?
B. What effect does the the depth at which the left mass is hung have on the condition of the system?
C. Is there any importance to the location of the points [tex]A[/tex] and [tex]B[/tex] (Given that [tex]l[/tex] remains the same)?
The book's answers (I'm sure about the book's answer for A, but the one and C looks like it was an answer to a different question, so I'd take it with a grain of salt.):
A. Yes.
B. None.
C. No.
My answers:
A+C. I got lost with the tension forces acting on A and B, so I took the easy way out.
I just calculated the center of mass, ignoring the various tensions as they are internal forces. It's on the axis of rotation, so the net moment on the fulcrum is 0, as long as the horizontal distance of both the masses is the same.
B. The lever arm for both of the identical forces of gravity acting on the masses is the same, so it doesn't matter.
It's just that it looks too easy, so I'd like some confirmation for the way I reached the answers.
Question 4
The force exerted by a stream of water on a rock on its bottom is proportional to the square of the speed of the current and the effective surface area of the rock according to the relation:
[tex]F=kAV^2[/tex]
It is given that at a certain point where the speed of the stream is [tex]V_1[/tex], rocks are dragged whose maximal weight is [tex]W_1[/tex]
At another point along the stream, the speed of the water is [tex]V_2=3V_1[/tex]
A. What is the maximal weight of the stones that are dragged at this point, assuming all the rocks have a similar shape (But different dimensions)
B. How will this result be affected if the rocks are cubic or spherical or of a different shape?
The book refers to an example where "Dragged by the stream" means overturned by it.
The book's answers:
A. The stream can carry rocks proportional to the velocity to the sixth power.
[tex]W_{max} \propto V^6[/tex]
So [tex]\frac{W_{2 max}}{W_1}=3^6=729[/tex]
B. The result is unaffected by the shape of the rocks.
Following the example, I looked at a cubic rock.
[tex]W=\rho a^3[/tex] where [tex]a[/tex] is the length of the side of the cube and [tex]\rho[/tex] is its density.
[tex]W \propto a^3[/tex]
The critical situation is when the normal force is about to leave the bottom of the rock. We'll demand that the net torque around the lower left corner of the rock is 0.
The moment of the contact reaction force is 0 as it acts from the lower left corner.
The moment of the force of gravity is [tex]\frac{Wa}{2}[/tex]
The moment of the force of the stream, which acts from the center of the side of the cube it's meeting, is [tex]\frac{Fa}{2}[/tex]
We'll demand that these two opposite moments are equal in magnitude.
The result is:
[tex]V^2=(\frac{\rho}{k})a[/tex]
[tex]V^2 \propto a \propto W^{1/3}[/tex]
[tex]V^6 \propto W[/tex]
Giving us the desired result.
However, this approach breaks down when considering a spherical rock, for which any force not acting through the line of action of gravity produces a net moment about the point of contact. I'd love if it someone could explain why the same result applies here.
Relevant image:
Well, I've got quite a few questions I couldn't solve. I'd love to be pointed in the right direction, because there are some problems I just don't know how to approach at all.
This is just the first batch, I'll be adding a couple more in a while.
I've attached a link to the relevant diagrams at the bottom of the post. Images removed pending approval.
Question 1
A fly is standing on the bottom of a close jar, positioned on one side of a sensitive balance scale which is initially leveled.
Suddenly, the fly rises up, hovers in place for several seconds, and touches back down.
Which of the following sentences most accurately describes what happens?
A. The scales remain leveled all the time.
B. The side with the jar will go down once the fly rises, stay leveled while the fly hovers in place, and come back up when the fly comes down.
C. The side with the jar will go up once the fly rises, stay leveled while the fly hovers in place, and come back when the fly comes down.
D. The side with the jar will go down all the while the fly is in the air.
The answer according to the book is B.
I am completely lost. Why would it go down if the fly is applying LESS force..? And why would it stay leveled if it's still imbalanced? Bleh, I can't make heads or tails of it.
Question 2
In the system in the diagram, the two masses, corresponding to [tex]W_1[/tex] and [tex]W_2[/tex] are given.
The coefficient of kinetic friction between the floor and the cart is [tex]\mu[/tex]
There is no friction between the two masses.
It is given that the cart is traveling with a constant velocity, [tex]V_1[/tex] to the right, and that the weight is descending at a constant velocity [tex]V_2[/tex]
A. Calculate the tension in the string.
B. Calculate the coefficient of friction, [tex]\mu[/tex]
C. What is the magnitude of the horizontal force which the cart exerts on the weight?
D. Calculate the ratio between the horizontal and vertical components of the weight's velocity, [tex]\frac{V_{2x}}{V_{2y}}[/tex]
E. How would the above results change if the coefficient of friction between the cart and the weight weren't 0?
My attempt:
The forces acting on the weight are:
Its weight, [tex]W_2[/tex]
The tension from the string, [tex]T_[/tex]
The normal force from the cart, [tex]N_{cw}[/tex]
Since it's moving at a constant velocity, then [tex]\Sigma \vec F=0[/tex]
Therefore:
[tex]N_{cw}=0[/tex] (This is the answer to C, and the book agrees with me on it.)
[tex]T=W_2[/tex] (This is the answer to A, and the book agrees with me on it)
The forces acting on the cart are:
Its weight, [tex]W_1[/tex]
The normal force from the floor, [tex]N[/tex]
The tension in the string, acting on the right of the cart, where its connected to it, [tex]T[/tex]
The kinetic friction, [tex]\mu N[/tex]
The tension in the string again, acting on the pulley on the top of the cart, [tex]T[/tex]
The normal force from the weight, [tex]N_{wc}=0[/tex]
Since it's moving at a constant velocity, then [tex]\Sigma \vec F=0[/tex]
Therefore:
[tex]N=W_1+T=W_1+W_2[/tex]
[tex]f_k=\mu(W_1+W_2)[/tex]
[tex]T=f_k[/tex]
Equating the two expressions for the the tension, I receive:
[tex]\mu=\frac{W_2}{W_1+W_2}[/tex]
The book's answer is:
[tex]\mu=\frac{2W_2}{W_1+W_2}[/tex]
Where did I go wrong so far? Should I have counted the tension a third time as well?
For every meter the weight descends, it pulls the string, and the cart, one meter to the right. So the ratio [tex]\frac{V_{2x}}{V_{2y}}[/tex] is 1.
The book's answer, however, is 1:2. I can't understand this.
Since the normal force between the weight and the cart is 0, any resulting friction would be 0, therefore, not changing the result by one bit.
Question 3
A uniform rod is placed horizontally at equilibrium on a smooth axis through its center. Two masses, [tex]m[/tex] are suspended on the rod as illustrated.
A. Will the rod remain horizontal? If not, which way will it turn?
B. What effect does the the depth at which the left mass is hung have on the condition of the system?
C. Is there any importance to the location of the points [tex]A[/tex] and [tex]B[/tex] (Given that [tex]l[/tex] remains the same)?
The book's answers (I'm sure about the book's answer for A, but the one and C looks like it was an answer to a different question, so I'd take it with a grain of salt.):
A. Yes.
B. None.
C. No.
My answers:
A+C. I got lost with the tension forces acting on A and B, so I took the easy way out.
I just calculated the center of mass, ignoring the various tensions as they are internal forces. It's on the axis of rotation, so the net moment on the fulcrum is 0, as long as the horizontal distance of both the masses is the same.
B. The lever arm for both of the identical forces of gravity acting on the masses is the same, so it doesn't matter.
It's just that it looks too easy, so I'd like some confirmation for the way I reached the answers.
Question 4
The force exerted by a stream of water on a rock on its bottom is proportional to the square of the speed of the current and the effective surface area of the rock according to the relation:
[tex]F=kAV^2[/tex]
It is given that at a certain point where the speed of the stream is [tex]V_1[/tex], rocks are dragged whose maximal weight is [tex]W_1[/tex]
At another point along the stream, the speed of the water is [tex]V_2=3V_1[/tex]
A. What is the maximal weight of the stones that are dragged at this point, assuming all the rocks have a similar shape (But different dimensions)
B. How will this result be affected if the rocks are cubic or spherical or of a different shape?
The book refers to an example where "Dragged by the stream" means overturned by it.
The book's answers:
A. The stream can carry rocks proportional to the velocity to the sixth power.
[tex]W_{max} \propto V^6[/tex]
So [tex]\frac{W_{2 max}}{W_1}=3^6=729[/tex]
B. The result is unaffected by the shape of the rocks.
Following the example, I looked at a cubic rock.
[tex]W=\rho a^3[/tex] where [tex]a[/tex] is the length of the side of the cube and [tex]\rho[/tex] is its density.
[tex]W \propto a^3[/tex]
The critical situation is when the normal force is about to leave the bottom of the rock. We'll demand that the net torque around the lower left corner of the rock is 0.
The moment of the contact reaction force is 0 as it acts from the lower left corner.
The moment of the force of gravity is [tex]\frac{Wa}{2}[/tex]
The moment of the force of the stream, which acts from the center of the side of the cube it's meeting, is [tex]\frac{Fa}{2}[/tex]
We'll demand that these two opposite moments are equal in magnitude.
The result is:
[tex]V^2=(\frac{\rho}{k})a[/tex]
[tex]V^2 \propto a \propto W^{1/3}[/tex]
[tex]V^6 \propto W[/tex]
Giving us the desired result.
However, this approach breaks down when considering a spherical rock, for which any force not acting through the line of action of gravity produces a net moment about the point of contact. I'd love if it someone could explain why the same result applies here.
Relevant image: