Question

A weight of 1.0 kg is suspended from the lower end of a wire of a cross section 10 millimetre square find the magnitude and direction of stress produced in it( 9.8 metre per second power
-2)

Answer 1

Considering...

Given -

m = 1kg

A = 10 mm2 = 10^-5 m2

g = 9.8 m/s2

F = ?

T = ?

Solution -

For suspended weight,

F = mg

F = 1×9.8

F = 9.8 N

Stress produced is given by,

T = F/A

T = 9.8/(10^-5)

T = 9.8 × 10^5 Pa

T = 980 kPa

Stress produced is 980 kPa in vertically upward direction.

Best luck

Answer 2

Here is your answer to the question.