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HINTS FOR PROBLEM SET #5
Problem Set #5 Problem 1 This is a problem of a system with two motions (block A block B)
and one degree of freedom.
(Since you can only move one of the blocks independently. i.e. the motion of A and B are dependent on each other. For an explanation of Degrees of Freedom of a System go to the bottom of this page.)
To get the relationship between the position of block A and the position of block B,
use the idea that the length of string is a constant.
This will give:
-xA + xB + xB + xB = constant (If you need more explanation to see how this equation was obtained, go to the hints for Problem Set #2 - Problem 3 and then go to the bottom of that page for an explanation of "The Conservation of String".)
Differentiating this relationship between the positions
gives the relationship between the velocities:
vA = 3vB
Differentiating again gives the relationship between the accelerations:
aA = 3aB Call this equation 1.
Now you will use Newton's Second Law (Fnet = ma) graphically.
Draw a FBD for block A.
You should do this with 2 colored pencils:
A bright green color for the body (block A) and a bright pink (not red) color for the forces acting on it. (Note: this will be a requirement for the FBDs on your exams.) Set this FBD equal to another diagram which can be thought of as the IVD (Inertia Vector Diagram).
The IVD shows the body with the inertia vectors acting on it.
Remember, an inertial vector is the acceleration vector times the mass of the body.
(Actually the inertial vector is the negative of the acceleration vector, so really, the IVD shows the body with the negative of the inertia vectors acting on it.) The IVD should also be in color.
A bright green color for the body (block A) and a bright pink (not red) color for the inertia vectors.
Now write Newton's second law for both the horizontal (x) and vertical (y) components.
The y component will give the value for the normal force acting on block A.
(Something we are actually not interested in for this problem.)
The x component will give an equation in T and aA.
Call this one equation 2.
Repeat the procedure for block B.
i.e. draw the FBD of block B and set it equal to the IVD for block B. The vertical component of Newton's Second Law will give an equation in T and aB. Call this equation 3.
3 equations in 3 unknowns: T, aA and aB You should be able to solve them.
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Problem Set #5 Problem 2 Part a)
You need to get the acceleration in the tangential direction for a toboggan just before it reaches a point B on the run. We'll call this point B-. Point B is a transition point between a curved portion of the run and a straight portion. Since you know the path of motion, use tangential and normal coordinates
to represent the acceleration vector of the toboggan.
(And besides that, you are asked to find the tangential component of the acceleration of the toboggan,
so you are forced to use these components.)
Draw the FBD of the toboggan at B- and then set it equal to the IVD at this point. Both these diagrams need to be in color. (See comment in Problem 1.) Apply the equation Fnet = ma in the two component directions. (tangential and normal) These equations allow you to solve for the acceleration in the tangential direction. You'll need the velocity at B- to calculate the normal component of the acceleration. an = (v2/ρ) This is not a problem since the velocity at point B is given.
Part b)
You need to get the acceleration in the tangential direction for the toboggan just after it reaches C. Call this point C+.
Do the exact same thing as you did at position B-. Using tangential and normal components to represent the acceleration vector,
draw the FBD and set it equal to the IVD. Apply the equation Fnet = ma in the two component directions. (tangential and normal) These equations allow you to solve for the acceleration in the tangential direction. You'll need the velocity at C+ to calculate the normal component of the acceleration, an = (v2/ρ) In this case you will run into a problem because the velocity at C+ is not given.
So you need to find the velocity at C+. You can assume it will be the same as the velocity at C-.
So you need to get the velocity at C-.
This is not as easy as it seems.
To do this you must analyze the motion on the straight portion of the curve, from B+ to C-.
Is this UARM? (Uniformly Accelerated Rectilinear Motion).
If it is, then you can use this equation:
vf2 = vo2 + 2ad where
vo is the velocity at B+ which is equal to the given velocity at B vf is the velocity of the toboggan at C- which is what you are looking for d is the distance between B and C
To check out if you have UARM,
draw the FBD/IVD for the toboggan in this straight portion of the curve.
Using Newton's second law you should find that indeed the acceleration is constant,
and furthermore, you should be able to find its value: a = 7.987 ft/sec2 (Note that this acceleration is not the same value you got for part (a) of this problem!) Plug this value of acceleration into the equation vf2 = vo2 + 2ad and solve for vf (which will be the velocity at point C+). Now use this value of the velocity at C+ in the equations you got for the FBD/IVD at C+.
This will give you the tangential acceleration at C+.
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Problem Set #5 Problem 3 If the collar D does not slide on the rod AB, then the path of motion of the collar is a circle in the horizontal plane with radius r and center on axis AC. Hence, the coordinate system of choice is one using tangential and normal components. This is a three dimensional problem and so you will also need to include the binormal direction.
The tangential direction is the direction tangent to the circle of motion. Show this direction as perpendicular to your paper, coming directly out of the paper towards you. The normal direction is the direction pointing towards the center of the circle. Show it in the plane of your paper and as a horizontal line directed towards the axis AC. The binormal direction is the direction upwards. Show it in the plane of your paper and as a vertical line directed upwards.
Draw the FBD of the collar. There are only two forces acting on it: The normal force from the rod and the weight of the collar. (You are told that friction is negligible.)
The force of the weight has only a component in the binormal direction.
The normal force is in the plane perpendicular to the axis of the rod. Break this up into two components: 1. Nt in the tangential direction, coming out of the paper directed at you.
2. Call the other N1 Note that N1 has components in both the normal and the binormal direction. N1 = Nnen + Nbeb Where Nn = N1cos40o Nb = N1sin40o So the entire Normal force vector is N = Ntet + Nnen + Nbeb = Ntet + N1cos40oen + N1sin40oeb
Beware! Don't get confused by the fact that the word "normal" is being used in two different ways in this problem. There is the normal force and there is the normal direction. They have nothing to do with each other! The normal force is the force normal to the axis of the rod. The normal direction is defined by the path of motion. It just so happens that in this problem the normal force has a component in the normal direction. (It also happens to have components in the tangential and binormal directions.)
Now draw the IVD of the collar. The acceleration vector will only have components in the tangential and normal directions. (The binormal component of acceleration is always zero.)
Set the FBD equal to the IVD and apply Fnet = ma. You only need to get the equations for the normal and the binormal directions. (Applying Fnet = ma to the tangential direction will not get you anything useful.) These two equations will allow you to solve for r.
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Problem Set #5 Problem 4 Assume that the block E does not slide in the slot.
Hence, its path of motion will be a horizontal circle of radius r about the axis AD where r = (26 10sinq) inches. (Dont forget to convert this to feet: 12 inches = 1 foot)
Hence, the coordinate system of choice is one using tangential and normal components. This is a three dimensional problem and so you will also need to include the binormal direction.
The tangential direction is the direction tangent to the circle of motion. Show this direction as perpendicular to your paper, coming directly out of the paper towards you. The normal direction is the direction pointing towards the center of the circle. Show it in the plane of your paper and as a horizontal line directed towards the axis AD. The binormal direction is the direction upwards. Show it in the plane of your paper and as a vertical line directed upwards.
Draw the FBD of the block E. It will have three forces: the weight of the block, a frictional force, and a normal force.
The force of the weight will be in the binormal direction only.
Since motion (or in this case, the tendency towards motion) is in the direction of the slot, the force of friction, which must always oppose motion, will be directed tangent to the slot. It will have components in the normal and binormal directions. It will not have a component in the tangential direction.
Note the word tangent is used here twice but in two completely separate ways. The term "tangent to the slot" is only referring to the fact that this it the direction the block wants to move and friction always opposes motion, or the tendency of motion. The tangential direction is determined by the path of
motion.
The normal force is in the plane perpendicular to the slot. Break this up into two components: 1. Nt in the tangential direction, coming out of the paper directed at you.
2. Call the other N1 Note that N1 has components in both the normal and the binormal direction. N1 = Nnen + Nbeb Where Nn = N1sinq Nb = N1cosq So the entire Normal force vector is N = Ntet + Nnen + Nbeb = Ntet + N1sinqen + N1cosqeb
Now draw the IVD of the collar. The acceleration vector will only have components in the tangential and normal directions. The binormal component of acceleration is always zero.
Set the FBD equal to the IVD and apply Fnet = ma. You only need to do this for the binormal and the normal directions. (Applying Fnet = ma to the tangential direction will not get you anything useful.)
These two equations will allow you to solve for the ratio of f/N in terms of q.
If this ratio f/N is less than the coefficient of static friction, then our assumption is OK. (Go back to the start, we assumed that the block E did
not slide in the slot.
Part a) Plug in q = 80o into the expression for f/N. You should get f/N = 0.1868 This is less than the coefficient of static friction and so our assumption is OK. So for q = 80o the block WILL NOT slide in the slot.
Part b) Plug in q = 40o into the expression for f/N. You should get f/N = 1.2123 This is more than the coefficient of static friction and so our assumption is not OK. So for q = 40o the block WILL slide in the slot.
Degrees of Freedom of a System - Explained The number of degrees of freedom that a system has is the number of moving parts that can be moved INDEPENDENTLY. This means the number of parts that you can arbitrarily chose the motion of. And once the motion of that number of parts has been chosen, the motion of any other moving part in the system is then determined.
In the example with three masses interconnected there are two degrees of freedom because you can move only two of the three masses independently (i.e. arbitrarily). In that particular case, the way the system is set up will allow you to arbitrarily move two of the masses in any way you chose. Once two of the masses have given motions, the third is determined.
It is true that if you only move one mass the other two will move as well, but you have no way of knowing exactly how they will move. i.e. there are several possibilities. Their motions are not completely determined. But if you move two of the masses in a specific (and arbitrary) way the third is determined.
The issue of degrees of freedom comes up in situations other than pulleys. For example gears. If you know what a planetary gear mechanism is, you should see that it has four motions and two degrees of freedom.
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