A scaffold of mass 50kg and length 5m is supported in a horizontal position by a vertical cable at each end. A window cleaner of mass 70kg stands at a point 2m from one end. Determine the tension in the nearer cable and also the farther cable.
The sum of the two cable tensions is
T1 + T2 = (M1 + M2)g = 1176 Newtons
M1 = board mass (50 kg)
M2 = cleaner's mass (70 kg)
For the tension T1,set the moment about the other end equal to zero. Let the cleaner be 2 m from cable 1. Take moments about cable 2, 3m from the man
T1*5 = M1*g*2.5 + M2*g*3
T1 = (1/5)*(1225 + 2058) = 657 N
T2 = 1176 - 657 = 519 N
To determine the tension in both cables, we need to consider the forces acting on the scaffold.
First, let's calculate the total weight of the scaffold and the window cleaner. The weight is equal to the mass multiplied by the acceleration due to gravity (9.8 m/s²).
Weight of the scaffold = mass × gravitational acceleration = 50 kg × 9.8 m/s² = 490 N
Weight of the window cleaner = mass × gravitational acceleration = 70 kg × 9.8 m/s² = 686 N
Next, let's analyze the forces acting on the scaffold in the horizontal direction:
1. The tension in the nearer cable pulls the scaffold to the left.
2. The tension in the farther cable pulls the scaffold to the right.
3. There is no net force acting on the scaffold in the horizontal direction because it is not accelerating horizontally (assuming no wind or additional external forces).
Since there is no horizontal acceleration, the sum of the horizontal forces must be zero:
Tension in the nearer cable - Tension in the farther cable = 0
We also need to consider the vertical forces acting on the scaffold:
1. The weight of the scaffold and the window cleaner pull the scaffold downward.
2. The vertical component of the tension in the nearer cable counteracts some of the downward force.
3. The vertical component of the tension in the farther cable also counteracts some of the downward force.
Since the scaffold is in equilibrium (not accelerating vertically), the sum of the vertical forces must be zero:
Weight of the scaffold + Weight of the window cleaner = Vertical component of tension in the nearer cable + Vertical component of tension in the farther cable
To calculate the tension in each cable, we need to determine the vertical component of the tension in each case.
In the nearer cable, the vertical component of the tension can be found using trigonometry (since it makes an angle with the vertical line). The angle can be determined using the right triangle formed by the scaffold length (5m) and the distance of the window cleaner from one end (2m). The angle can be found using the inverse tangent function:
Angle = atan(vertical distance / horizontal distance) = atan(2m / 5m)
Once we have the angle, we can find the vertical component of the tension in the nearer cable using the sine function:
Vertical component of tension in the nearer cable = Tension in the nearer cable × sin(angle)
Similarly, we can calculate the vertical component of the tension in the farther cable:
Vertical component of tension in the farther cable = Tension in the farther cable × sin(angle)
Since the total weight of the scaffold and the window cleaner is equal to the sum of the vertical components of the tensions, we can set up the following equation:
Weight of the scaffold + Weight of the window cleaner = Vertical component of tension in the nearer cable + Vertical component of tension in the farther cable
Now we have two equations:
Tension in the nearer cable - Tension in the farther cable = 0
Weight of the scaffold + Weight of the window cleaner = Vertical component of tension in the nearer cable + Vertical component of tension in the farther cable
Using these equations, you can solve for the tension in both cables simultaneously by rearranging the equations and substituting the appropriate values. With the given information, you should be able to calculate the tension in the nearer cable and the farther cable.