A 78.0 kg person is being pulled away from a burning building as shown in Figure 4.27. Calculate the tension in the first rope, T1, if the person is momentarily motionless.
T1 = 736 N
To calculate the tension in the first rope (T1), we can use Newton's second law of motion, which states that the sum of the forces acting on an object is equal to the mass of the object multiplied by its acceleration.
In this case, the person is momentarily motionless, so their acceleration is zero. This means that the sum of the forces acting on the person must also be zero.
The forces acting on the person are the force of gravity (mg, where m is the mass of the person and g is the acceleration due to gravity) and the tension in the first rope (T1).
Since the person is motionless, the tension in the first rope must balance out the force of gravity:
T1 = mg
Given:
Mass of the person (m) = 78.0 kg
Acceleration due to gravity (g) = 9.8 m/s^2
To calculate T1, we can substitute the values into the equation:
T1 = (78.0 kg)(9.8 m/s^2)
Calculating this, we get:
T1 = 764.4 N
Therefore, the tension in the first rope (T1) is 764.4 Newtons.
To calculate the tension in the first rope, T1, when the person is momentarily motionless, we need to consider the forces acting on the person.
Let's assume that the person is connected to two ropes, T1 and T2. T1 is the tension in the first rope and T2 is the tension in the second rope.
When the person is motionless, the forces acting on the person are balanced, which means that the net force on the person is zero. The weight of the person, acting downwards, is balanced by the tension forces in the ropes.
The weight of the person can be calculated using the formula:
Weight = mass × acceleration due to gravity
Weight = 78.0 kg × 9.8 m/s²
Weight = 764.4 N
Since the person is motionless, the tension in the first rope, T1, will be equal to the weight of the person.
Therefore, the tension in the first rope, T1, is 764.4 N.