A 3.0 m long light board with negligible weight is supported at each end by cables. a painter weighing 900 N stands 1.o m from the left cable. Find the tension in each cable.
900(1.0)+ Tright(3)
so
Tright = 300 N
Tright + Tleft = 900
so
Tleft = 600 N
To find the tension in each cable, we need to analyze the forces acting on the light board.
Let's label the left cable as Cable A and the right cable as Cable B.
First, let's consider the forces acting vertically (in the y-direction):
1. The weight of the painter (900 N) acts downward at a distance of 1.0 m from Cable A.
2. The weight of the light board is negligible and can be ignored.
Next, let's consider the forces acting horizontally (in the x-direction):
1. There are no horizontal forces acting on the system because the light board is weightless.
To maintain equilibrium, the sum of forces in the x-direction and y-direction must be zero.
Now, let's analyze the forces acting vertically (in the y-direction):
1. The weight of the painter (900 N) creates a downward force that needs to be balanced by the upward forces from both cables.
2. Cable A exerts an upward tension force.
3. Cable B exerts an upward tension force.
Since the light board is at rest, the vertical forces must be in equilibrium.
Let's calculate the tension in each cable:
We can start by considering the vertical forces balancing:
900 N (weight of the painter) = Tension in Cable A + Tension in Cable B
Since the light board is distributed evenly and there are no other vertical forces acting on it, each cable will support half of the weight of the painter:
Tension in Cable A = Tension in Cable B = (900 N) / 2 = 450 N
So, the tension in each cable is 450 N.