Question

24. A homogeneous chain lies in limiting equilibrium on a horizontal table of coefficient of
friction 0.5 with part of it hanging over the edge of the table. The fractional length of the
chain hanging down the edge of the table is​

Answer 1

Answer:

This is fairly an easy question, if you understand the concept of Mass Gradient, which in itself is natural and intuitive. I will show the working of the solution for the problem in following steps:

Step 1: Assume total length of the chain to be = L. The length of the portion hanging = x. The remaining part = L-x

Step 2: Since the chain is uniform, its mass gradient is constant throughout its length. Its mass gradient = M/L.

Step 3: The gravitational force on the hanging part = mg = (M/L)xg. This is the force acting on the part of the chain resting on the table top. Draw its free-body diagram, showing all the forces acting on it, including Normal force and Frictional force.

Step 4: Since the gravitational force and normal force cancel out each other, the body stays in equilibrium along Y-axis. All you must check for is the condition for horizontal equilbrium. For this, the force acting on resting part due to hanging part, and the frictional force must balance each other. This is the major part of the problem done, with just some analytical thinking.

Step 5: Now comes the working part,

(M/L)xg = k (M/L)(L-x)g

i.e., Force due to hanging part = Frictional force

Now, value of frictional constant k is given to be 0.25, which would give us-

x = (k/k+1)L

→ x = L/5

Therefore, one-fifth of the chain would be hanging from the table.

Now, I would like to share some views on the critical concepts that we used in the above problem:

→ Whenever you have a problem on chains, rods,etc., i.e., objects having their linear mass densities as a constant, always use the concept of Mass Gradient. It would make the problem look much easier.

→ Be careful when you draw free body diagrams of the extended objects. Always show that gravitational force acts on the Centre of Mass of the body and that frictional force acts along the surfaces in contact. This would help you much when you are also considering the Torques due to different forces.

I sincerely hope that the above answer would elucidate the concept for you. If you have any other queries regarding the above example, you are always welcome to ask!

Explanation:

please amrkme as brainliest

Answer 2

Answer:

ANSWER IS 1/3