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Two objects, each of weight W, hang vertically by spring scales as shown in the figure. The pulleys and the strings attached to the objects have negligible weight, and there is no appreciable friction in the pulleys. The reading in each scale is

A: more than 2W.
B: less than W.
C: more than W, but not quite twice as much.
D: W.
E: 2W.

Two objects, each of weight W, hang vertically by spring scales as shown in the figure-example-1

2 Answers

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Final answer:

In the provided physical setup with negligible weight pulleys and strings, and no friction, the spring scale reading for each object of weight W is simply W. This is because each scale measures the direct weight of the object hanging from it. The correct option is (D).

Step-by-step explanation:

Understanding Spring Scales in Physics

When analyzing the scenario where two objects of weight W each hang from spring scales, we must apply the principles of statics and equilibrium.

Since the pulleys and strings have negligible weight and there is no appreciable friction, each scale will only measure the weight of the object hanging from it.

Therefore, the reading on each scale will be W, as the scales are in equilibrium with the downward force of weight W being balanced by the upward force exerted by the spring scale.

Considering other options such as more than 2W, less than W, or more than W but not quite twice as much, we dismiss these because there are no additional forces acting on the scales that would cause such readings.

Each object simply exerts a force equal to its weight on the corresponding scale, and this is the force that the scale measures.

Now, addressing some of the other contexts provided for reference:

  • To weigh an object whose weight is larger than the maximum reading of a scale, a simple fulcrum and leverage can be used to proportionately reduce the force applied to the scale.
  • The likelihood of a ladder slipping depends on the angle and the friction between the ladder and the surface; the risk is greater if these factors are not properly accounted for, regardless of the position of the painter on the ladder.
  • For two springs with different force constants, the one with a greater force constant (Spring A) will exhibit less extension than a softer spring (Spring B) when equal weights are suspended from both.
  • When considering pendula with different mass bobs displaced by the same angle, the period of the pendulum is independent of the mass of the bob and both pendulums will swing with the same frequency if their lengths are equal.

User Sital
by
8.7k points
3 votes

Answer:

D: W.

Step-by-step explanation:

For an ideal spring connected to weight W

Now here since both sides we have same weight connected so the system will always remains at rest

So here we have


W - W = ma


a = 0

so here acceleration will be zero for the given system

now we can find the tension force in the string on each side as


T - W = ma


T = W

so here tension on string on side will be same and it is equal to W

so the reading will of the scale is W

User Stephan Ronald
by
8.4k points