## Sunday, January 1, 2017

### Mechanical Properties of Solids Problems with Solutions Five

We are solving series of problems on mechanical properties of solids. It is the nature of the solid material that they can recover to its original position up to some extend and it depends on its nature. When a longitudinal stress is applied on the wire, there is a linear extension and simultaneously there is lateral contraction along the thickness off the wire. The ratio of lateral contraction strain  to linear extension strain  is called poission's ratio. For a perfectly elastic body its value is half and it has no units and dimensions. No body can have Poission's ratio greater than half and in practice it is less than half. We need to do some work in producing elongation in the wire and that work done is stored in the form of elastic potential energy.

Problem

A steel wire of known diameter, cross section and length is clamped at  two points and the separation between them is given to us as shown in the diagram below. A body of known mass is suspended from the wire at the middle point and the sag is also given to us. If Young's modulus of the wire is given to us, we need to know the mass that is suspended.

Solution

We can  resolve the tension in the wire as components and the part of the tension can be balanced by the weight at the equilibrium condition as shown in the diagram below. We can  find the tension value from the given  data and solve the problem as shown in the diagram below.

Problem

A stone of known mass is attached at one end of aluminium wire of known dimensions and is suspended vertically. The stone is now rotated in horizontal plane at a known angle as given in the problem below. We need to find the increase in the length of the wire.

Solution

We can resolve the tension in the wire into components and we can equate the part to the weight of the body  as shown in the diagram below. By substituting this tension as the force in the definition of the Young's modulus, we can solve the problem as shown in the diagram below.

Problem

A light rod of length is suspended from a support horizontally as shown in the diagram below with two vertical wires of equal lengths and we have information regarding their modulus and area of cross section. We need to  know where we need to suspend a weight so that the stress in both the wires are equal and the problem is as  shown in the diagram below.

Solution

We know that stress means force per unit area. Force in the wire nothing but the tension in the wire. The system has to be in equilibrium and for that the torque about a point has to be balanced and the problem is solved as shown in the diagram below.

Problem

Two blocks of known masses are connected through a wire of breaking stress known over a friction less pulley and we need to know the maximum radius of the wire used so that the wire is not going to break and the problem is as shown in the diagram below.

Solution

We can find the stress in the wire as the ratio of force per unit area. We can write equation for the resultant force as the difference between the tension and the weight respectively as shown in the diagram below. By solving both the equations, we can find the tension as well as the radius of the wire as shown in the diagram below.

Problem

Combination of steel and brass  rods of known area of cross section are joined together to support a load as shown in the diagram below. We need to find the stress in each rod and the problem is as shown in the diagram below.

Solution

As the system is horizontal decrease in the length of both the wires has to be the same. By equating them we can find the relation between the stress of both the wire and by equating it to the weight given in the problem we can solve the problem as show in the diagram below.

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