Question

a uniform rod length 6m and mass1 = 42 kg is hinged at A, makes angle of 35 degrees below horizontal, and is supported by a tungsten cable attached to a 1.5 m from other end. The cable makes an angle 20 degrees above horizontal and with a load of mass2 = 80kg is hanging at the end.

A. Find tension in cable (horizontal and vertical) at the pivot point hinge A.

B. max tension is 2800 N; find max distance from A at which cable can be reattached before cable breaks

C. Initial tungsten cable legth is 7.2 m before theres any tension, cable diameter is 4.3 mm; Young's modulus is 35 x 10^10 N/m^2, if cable is attached at 1520 N, find amount cable is stretched

Answer 1

A ) The formula for horizontal tension is given by

T = M1gCos1-------(1)

here M1 = 42Kg

     1 = 35^0

substituting in eqn (1) we have

T = (42x10^-3)(9.8)Cos(35)

T = 337.14x10^-3N ALONG THE HORIZONTAL

The formula for vertical tension is given by

T' = M2gSin2---------(2)

here M2 =   80Kg

2 = 20^03

substituting the values in eqn(2) we have

T' = (80x10^-3)(9.8)Sin(20)

T' = 268.12x10^-3N ALONG THE VERTICAL

C)length l = 7.2m

diameter d = 4.3mm = 4.3x10^-3m

young's modulus Y = 35x10^10N/m^2

force F = 1520N

HERE radius r = d /2 = (4.3x10^-3)/2

   r = 2.15x10^-3m

area A = r^2

          =(3.14)(2.15x10^-3)^2

A = 14.5x10^-6m^2

the formula for young's modulus is given by

Y = STRESS/ STRAIN= F L / A L-----(3)

SUBSTITUTING THE VALUES IN EQN (3) WE HAVE

3.5x10^11 =(1520)(7.2) / (14.5x10^-6)(L )

SO

L = 2.15x10^-3m This is the amount at which the cable is stretched