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Mechanics
Physics: Problem of the day-17
Jun 7th
Topic: Mechanics
Threads of lengths h1, h2 and h3 are fastened to the vertices of a homogeneous triangular plate of weight W. The other ends of the threads are fastened to a common point, as shown in the figure below.
Figure-1
What is the tension in each thread, expressed in terms of the lengths of the threads and the weight of the plate?
Clearly, the centre of mass S of the triangle has to be below the point of suspension. Denote the vectors pointing from the centre of mass S to the vertices of the triangle by r1, r2 and r3, and that to the suspension point by m (see figure).
Figure-2
The forces F1, F2 and F3 exerted on the plate by the threads can now be expressed in terms of the vectors defining the thread directions:
Fi = λi (m – ri), i = 1, 2, 3
Since the plate is in equilibrium, the vector sum of the forces acting on it is zero, i.e.
F1 + F2 + F3 + W = 0
Making use of the fact that the vector pointing to the center of mass of the triangle (the origin of the vector reference frame) is the arithmetic mean of the vectors pointing to its vertices, we have
r1 + r2 + r3 = 0
We note that W and m are parallel and, therefore, W = –km. Eliminating r3 from the above equations gives
(λ3 – λ1) r1 + (λ3 – λ2) r2 + (λ1 + λ2 + λ3 – k) m = 0
Since r1, r2 and m, are not in the same plane, a linear combination of them can only be zero if the coefficient of each is zero.
Thus, λ1 = λ2 = λ3 which implies that the tensions in the threads are proportional to their lengths. This deduction would become invalid if one of the threads were slack, since the plane of the plate would then become vertical and m would lie in it.
Physics: Problem of the day – 16
May 17th
TOPIC : Mechanics
Q. Three loads of mass m1, m2 and M are suspended on a string passed through two pulleys as shown:
Figure-1
The pulleys are at the same distance from the point of suspension. What is the ratio of these masses at which the system is in equilibrium?
The easiest way to reach a solution is to consider the point at which the three strings join each other:
Figure-2
Since this point is in equilibrium,
which means that the three vectors should be able to form at triangle:
Figure-3
Therefore, we only require that:
T1 + T2 > T3, T2 + T3 > T1, T3 + T1 > T2
=> m1 + m2 > M, m2 + M > m1, M + m1 > m2
If these conditions hold, the system can attain equilibrium.
The same conditions can be derived more analytically by considering the angles the two inclined strings form with the horizontal, and writing the force equations in more details – that is an exercise for you.
Physics: Problem of the day – 15
May 11th
Topic: Mechanics
The masses of two stars are m1 and m2 their separation is l. If this star-system is to remain stable, the stars must be revolving about the system’s center of mass, with the force of gravity providing the necessary centripetal force. From the information provided, is it possible to calculate the time period of the stars revolution?
Figure-1
Let X be the center of mass of the system. Obviously,
Let the common angular velocity of the stars about X be ω.
The time period T is inversely dependent on ω:
Therefore, the information we were provided about the star-system was sufficient to calculate the time period of revolutions of the stars about the system’s center of mass.
It might seem to you that we did not even use one of the equations, namely m1l1 = m2 l2 , which means we have one extra equation than required. Is that true? If it is, we could have determined the time period using only two of the three constants provided to us.
Physics: Problem of the day – 13
May 8th
Topic: Mechanics
Q. A cylindrical tube of length L which has inside it a light spring of length l and spring constant k, is placed vertically on a table. A small ball of mass m is released at the top of the tube.
Figure-1
At what height h from the surface of the table does the ball have maximum velocity?
For a reader who is alert enough, this problem is a matter of seconds. We just have to visualize what happens to the ball as it falls. It accelerates until it has reached the top of the spring. Now, as it compresses the spring, its acceleration will decrease due to the spring whose compression causes it to exert a progressively increasing force on the ball. But it will nonetheless accelerate, up to the time when it can no longer have a downwards acceleration because the spring has been highly compressed. At this point, the spring force on the ball balances its weight, and the ball has attained its maximum velocity. After this of course, the ball starts slowing down.
Therefore, at h,
Note that L was of no use to us!
Physics: Problem of the day – 11
May 3rd
This problem is again a challenge for you.
Q: Suppose that a body moves in a medium which offers drag to its motion, the drag force F being proportional to the velocity v of the body, raised to some power b, that is, F = avb
If the body is imparted an initial velocity, is it possible for it to cover an infinite distance?
Let us rephrase this question a bit. You have to figure out whether there is any value of b possible for which if the body is given an initial non-zero velocity, it never stops.
The question might seem strange. After all, if there is drag on the body, its velocity should be reduced continuously and finally become zero in some finite amount of time, so it should eventually stop. Well, there is no doubt about the truth of the statement “Its velocity should be reduced continuously”. But the second part, “The velocity finally becomes zero in some finite amount of time”, might not be very true. This is what you have to think about.