A 1.5 kg mass is held at rest on top of a frictionless and horizontal table. A l
ID: 1906201 • Letter: A
Question
A 1.5 kg mass is held at rest on top of a frictionless and horizontal table. A light string loops over a pulley which is in the shape of a 10 cm radius solid disk which has a mass of 1.5 kg. The light string then supports a mass of 1.5 kg which is hanging in air.
A) The mass on the table is released and the suspended mass falls. What is the acceleration of the falling mass.
B) What is the tension in the string which is attached to the sliding mass on the table?
C) What is the tension of the string which supports the hanging mass?
Explanation / Answer
Depends on the configuration of the string system. The key to any string configuration is to recognize that a fixed string and whatever is attached to it are not moving. This means that the sum of all forces acting on that string and attachments is zero (i.e., f = 0; where f is the net force acting on the string). If f were not zero (e.g., f > 0), the string and everything else attached to it, like a weight (W), would be accelerating. This stems from f = ma; where m is the mass of the system and a is its acceleration. You can see from this that the only way the string system can remain static is for f = ma = 0; so that a = 0 and the system is not accelerating. So to find that "equation": First, identify ALL the forces acting on the system (string plus whatever else is attached to it, like a weight). Then add them up as vectors (magnitude with direction). If the sum you get (f) is not zero (e.g., f0), then you've overlooked a force or did the math wrong. This follows from f = ma = 0 for a non-accelerating string system. Second, identify the force vector that follows the length of the string. That's the so-called tension. Understand, it's tension if and only if the string is tense (taut). If it's flopping about, there is no tension per se. Third, divide all the forces into their X and Y components; e.g., F = Fx + Fy as vectors. Fourth, write two equations, which are in general: Sum of Fx to the left = Sum of Fx to the right Sum of Fy up = Sum of Fy down We know all the forces in the X directions (left and right) add up to zero (are equal but opposite) and all the forces in the Y directions (up and down) also add up to zero. Why? Because that string system is static, it is not moving in any direction; so the net forces up and down, and left and right have to be zero as well. In fact, this is what we really mean when we say f = ma = 0; that is, all the forces (up, down, left, right) add up to zero and give the system a = 0 acceleration. Fifth, solve the two equations for the tension force.Related Questions
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