Steven W. answered • 07/08/18
Tutor
4.9
(4,291)
Physics Ph.D., college instructor (calc- and algebra-based)
Hi Achillos!
I will assume these problems are numbered 1 - 5, top to bottom.
1. This problem uses the equilibrium conditions from statics. Using x-y (Cartesian) coordinates, I usually write these as three conditions:
(a) sum of the x-direction forces = 0 (ΣFx = 0)
(b) sum of the y-direction forces = 0 (ΣFy = 0)
(c) sum of the torques = 0 (Στ = 0)
EQUILIBRIUM ANGLE OF THE STRING (using the sum of forces equations)
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I drew up a little diagram of the situation, as I understood it, to refer to. The only assumptions I made were that the meter stick was uniform, and the angles desired were between the string and the vertical, and the meter stick at the vertical:
https://imgur.com/qq01Cw9
From this, you may be able to see that the only x-direction forces are the horizontal force on the end of the meter stick (which is known) and the horizontal component of the tension, which you can write in terms of the overall tension (T) and the angle (θ) between the string and the vertical (see if you can figure out how to do that, given the drawing).
The only y-direction forces are the force of gravity on the meter stick (known), and the y-component of the tension, which can be written again in terms of T and θ.
With this information, you can fill in expressions for all the terms in the static equilibrium equations (a) and (b) above. Once you do that, you will have two equations with two unknowns: T and θ. By solving these equations as a system, you should be able to solve for the angle θ, which I took to be the "equilibrium angle of the string."
EQUILIBRIUM ANGLE OF THE METER STICK (sum of torques equation)
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We can use the sum of torques equation to get the angle I called α in the diagram, the "equilibrium angle of the meter stick." I would recommend using the point where the string attaches as the center of rotation. This way, the tension exerts no torque and can be left out of the sum of torques equation.
You can then look at the torques created by the other two forces (gravity and the pushing force) around that axis where the string attaches to the meter stick. To determine the torque exerted by a force around a point, you have to know three things:
1) the force
2) the distance from the chosen axis of rotation to where the force is applied (which I usually call "r")
3) the angle (which I will call γ to avoid confusion with the angles listed above) between the direction of the force and the direction of r
Then the torque of that force around that center is: τ = Frsin(γ)
Choosing the axis of rotation at the point where the string attaches, and knowing that gravity is applied at the center of mass (which, if the meter stick is uniform, is at the geometric center of the stick), and the pushing force at the end, should allow you to write torque expressions for gravity and the pushing force around that axis. You should find that the γ angle between the force and r in each case should depend on α, the angle between the meter stick and the vertical (I drew the r's on the diagram). This makes α the only unknown in the sum of torques expression, so you should be able to solve for it.
I may have said some things here you already know, but I hope these ideas set you on the right track for this one. I will be back in a little while to make notes on the other ones, as well. If you have any questions, or would like to check an answer, please do not hesitate to ask.
Steven W.
tutor
3. For this one, refer to this diagram I made:
https://imgur.com/GL8n39T
In this case, the three forces acting on the car are the normal force of the incline, gravity, and the maximum force of static friction on the track. Because we are looking for the maximum permissible speed without skidding, we assume not only that the friction is the static kind (no sliding) and at its maximum value (fsmax = μsN), but also that it points down the plane, to keep the car from skidding toward the outside of the turn if it goes to fast (we can show why it must point down the plane in this case, if necessary, but I will leave it as a given for now).
The result of these forces is that the car stays in a horizontal circle, and does not move vertically. This means the net force on the car should be 0 in the vertical (y) direction, and the centripetal force (Fc) in the horizontal direction. From the diagram, you may be able to see that:
ΣFx = Nx+fsmax-x = Fc = mv2/r (where v is the maximum permissible speed of the vehicle, and r is the radius of the curve)
ΣFy = Ny + fsmax-y - mg = 0
Since fsmax = μsN, these two equations -- once appropriate expressions are put in, will have only two unknowns, N and v. You can then solve this system of equations for v.
See if you can use these comments to get yourself started on this one, and if you have any other questions or problems, just let me know! I will be back in a little while with thoughts on the other two.
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07/08/18
Steven W.
tutor
4. For this one, about the hanging plumb at some latitude, you can refer this diagram:
https://imgur.com/Hm4VoOe
Normally, we consider a plumb bob to hanging directly toward the center of the Earth, countering gravity. However, if we consider the Earth's rotation, there also has to be an acceleration at any position that is perpendicular toward the Earth's axis. This is because there must be a centripetal acceleration in that direction, associated with traveling in the dashed-line circle I drew on the diagram.
That centripetal acceleration is given by:
ac = v2/r
where v = speed of the point on the circle
r = radius of the circle
From the drawing, you may be able to see that, using geometry and trigonometry, you can write an expression for the radius of the dashed-line circle (which I called RL) in terms of given quantities.
Once you have the radius, you need to get v. Assuming the Earth is rotating at constant speed (a good assumption), you can use:
d = vt (distance = speed * time; the constant-speed equation of motion)
For one full traversal of the circumference of the circle, the distance d = 2πRL, and the time it takes will be one day (which you can convert to seconds. You can then calculate v (in m/s).
Knowing v and RL, you can then calculate the centripetal acceleration (ac) necessary to keep something in that location moving in that dashed line circle.
The only acceleration that points in the correct direction to do that hear is g (the acceleration of gravity). If g = 9.81 m/s^2 overall, and its component (which I called gL) along the radius of the dashed circle has to equal ac, then, using trigonometry, you should be able to calculate the angle θ between g and what gL, perhaps using the triangle I listed in the drawing. This will get you θ in terms of the latitude angle α, and you can then compute a value for θ at α = 45°.
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07/08/18
Steven W.
tutor
I apologize, Achillios! I messed up #4 about the Earth. Here is a new image that is corrected:
https://imgur.com/FO9sVkr
https://imgur.com/FO9sVkr
The actual deviation angle off of the radial line is indicated as θ. It can be obtained by setting up the static equilibrium sum of forces equations in both x and y, as we have done before. In the y direction, the sum should be zero, because there should be no resultant force. In the x direction, the resultant force will be the centripetal force, whose acceleration I already described in my previous comment about this problem. So we have:
ΣFx = Tcosβ - mgcosα = -mac (the negative sign in front of the term on the right side of the equation is because I defined the inward direction in x as negative, and that is the way the centripetal force points)
ΣFx = Tcosβ - mgcosα = -mac (the negative sign in front of the term on the right side of the equation is because I defined the inward direction in x as negative, and that is the way the centripetal force points)
ΣFy = Tsinβ - mgsinα = 0
The key here is that we are allowed to write the answer in terms of α, so we can treat that as a "known." So the only unknowns in these equations are T and β. As before, we can then solve these as a system to get an expression for β (in terms of α and other known quantities).
Once we have an expression for β, we can get an expression for θ = (β - α). The angle θ (as shown in the inset of the above-linked image, is the angle by which the plumb string deviates from an extension of the radial line.
I apologize for the error and any further confusion this may have caused. If you have any questions about any of these, please do not hesitate to ask!
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07/08/18
Steven W.
07/08/18