Given DIAGRAM 1.1 where FT means Tension and W means Weight, calculate the tension
on the steel cable given the following conditions:
A. The cargo is stationary.
B. The cargo accelerates upward at a rate of 0.25 m/(s^2)
355 kg cargo
Without seeing DIAGRAM 1.1, I will assume that it is a diagram of a cargo suspended by a steel cable. In both cases, the weight of the cargo (W) will be the same and can be calculated using the formula:
W = m * g
where m is the mass of the cargo and g is the acceleration due to gravity (approximately 9.8 m/s²).
Given that the mass of the cargo is 355 kg, we can calculate the weight:
W = 355 kg * 9.8 m/s²
W = 3481 N
A. When the cargo is stationary, the tension in the steel cable will be equal to the weight of the cargo. Therefore, the tension (FT) in the steel cable is 3481 N.
B. When the cargo accelerates upward at a rate of 0.25 m/s², there will be an additional force acting on the cargo in the upward direction. This force can be calculated using the formula:
F = m * a
where m is the mass of the cargo and a is the acceleration.
Given that the mass of the cargo is 355 kg and the acceleration is 0.25 m/s², we can calculate the additional force:
F = 355 kg * 0.25 m/s²
F = 88.75 N
The tension in the steel cable will be the sum of the weight of the cargo and the additional force:
FT = W + F
FT = 3481 N + 88.75 N
FT = 3569.75 N
Therefore, the tension in the steel cable when the cargo accelerates upward at a rate of 0.25 m/s² is approximately 3569.75 N.