Refer to the 2th Law of Newton
F = m. a
F = 15 x 2.8 = 42 N
Answer:
1 mi = 5280 ft * 12 in/ft = 63360 in
A convenient conversion factor (to remember) is 1 m = 39.37 in
63360 in / (39.37 in / m) = 1609.3 m
26 mi + 285 m = 26 * 1609.3 + 385 = 42,228 m
Answer:
The total elongation for the tension member is of 0.25mm
Explanation:
Assuming that material is under a linear deformation then the relation between the stress and the specific elongation is given as:
(1)
Where E is the modulus of elasticity, σ the stress and ε the specific deformation. Also, the total longitudinal elongation can be expressed as:
(2)
Here L is the member extension and δL the change total longitudinal elongation.
Now if the stress is found then the deformation can be calculated by solving the stress-deformation equation (1). The stress applied sigama is computed dividing the axial load P by the cross-sectional area A:
Solving for epsilon and replacing the calculated value for the stress and the value for the modulus of elasticity:




Finally introducing the specific deformation and the longitudinal extension in the equation of total elongation (2):