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Mechanics Interesting Help ...Newtons Laws?
One end of a light inextensible string is attached to a fixed point A of a fixed vertical wire. The other end of the string is attached to a small ring B, of mass 0.2 kg, through which the wire passes.
A horizontal force of magnitude 5 N is applied to the midpoint M of the string. The system is in equilibrium with the string taut, with B below A, and with angles ABM and BAM equal to 30 degrees (SEE DIAGRAM).
(i) Show that the tension in BM is 5N.
ANSWER: This part is OK. I used Triangle law of forces and checked it with Lami's theorem. The tension is in the string BM or MB. But then, what should be the direction of the tension force????? Should it be vector BM or vector MB???????????????? This is indeed very confusing.
(ii) The ring is on the point of sliding up the wire. Find the coefficient of friction between the ring and the wire.
ANSWER: We need to resolve vertically. But, if vector BM is considered, then, the y component of the tension is 5cos(30) and if vector MB is considered, then, the y component of the tension is - 5cos(30).
SO, where is my confusion?????????????????????????????
(iii) A particle of mass m kg is attached to the ring. The ring is now on the point of sliding down the wire. Given that the coefficient of friction between the ring and the wire is unchanged, find the value of m.
PLEASE share your valuable ideas and explain me my confusion about the direction of tension.
THANKS..
1 Answer
- AngelosLv 57 years ago
tension at a point of a taut string is the force that would be needed to keep the two parts of the string together if the string were cut at that point. Manifestly, at every point of the string other than at the ends the tension points in BOTH directions: each part of the string pulls on the other part (think of a rubber band!). At each end of the string, the tension points towards the string (again, think of a rubber band). In your example, the tension is along BM (i.e. towards M) at B and along MB (i.e. towards B) (and also along MA, i.e. towards A) at M -- which is why point M is at rest, with the three forces acting upon it being equal in magnitude and having zero resultant.
II. The ring is pulled upwards by the vertical component of the tension (5cos(30°) = 5√3/2 = 4.33 N, as you correctly say) and downward by its own weight Mg = 0.2*9.8 = 1.96 N. The two forces do NOT cancel (their resultant is 2.37 N pointing up), and the ring would slide upwards but for friction. However, it is also being pulled horizontally against the wire with a force of magnitude 5sin(30°) = 2.5 N (the horizontal component of the tension at B). Since the resulting friction is just barely enough to keep it in place, the coefficient of friction must be 2.37/2.5 = 0.948
III. Since the ring is now on the point of sliding down the wire, the weight of the ring, which is 1.96 N, plus the added weight mg of the particle, MINUS the upward-pulling component 4.33 N of the tension, must exactly counterbalance the maximum friction between the ring and the wire, which is still 2.37 N. So the weight mg of the particle is 2.37 + 4.33 - 1.96 = 4.74 N and its mass is 4.74/9.8 = 0.484 kg = 484 grams.
