Question

the bar ABC is supported by a pin A and a steel wire at B. calculate the elongation of the wire when the 36-lb horizontal force is applied at C. The cross sectional area of the wire is 0.0025 in² and the modulus of elasticity of steel is 29x10^6 psi.​

Answer 1

Answer:

the answer is 0.05

Step-by-step explanation:

deformation = PT/AE

deformation is Directly proportional to PL and inversely proportional to AE

1.) ∑M@A= 0

T= 36(10)/6

T= 60 lb

2.) Deformation = 60(5x12)/0.0025(29x$10^{6}$)

Deformation = 0.05in.

answer is 0.05 in

Answer 2

Answer:

The solution is 0.05

Step-by-step explanation:

'Deformation'  = $\frac{PT}{AE}$

'Deformation' is : Directly proportional to PL and inversely proportional to AE

1) ∑M@A= 0

T= $\frac{36(10)}{6}$

T= 60 lb

2.) Deformation = $\frac{60(5(12))}{0.0025(29(10^{6} ))}$

'Deformation' = 0.05in.

Therefore, Deformation = 0.05in.