A 4-m wooden platform weighing 160 N is suspended from the roof of a house by ropes attached to its ends. A painter weighing 640 N stands 1.2 m from the left-hand end of the platform. Find the tension in each of the ropes
Perform a total force and total moment (torque) balance.
The sum of the two rope tensions is 800 N.
Set the total moment about either end equal to zero to obtain the tension of the rope on the other side. The moment due to the weight of the plank acts like a 160-N weight at the center
Please solve for me
To find the tension in each of the ropes, we can set up and solve a system of equations based on the equilibrium of forces.
Considering the forces acting on the platform, we have:
1. The weight of the platform (160 N) acting downward.
2. The tension in the left rope (T1) acting upward.
3. The tension in the right rope (T2) acting upward.
4. The weight of the painter (640 N) acting downward.
Considering the torques acting on the platform, we have:
1. The torque due to the weight of the platform (160 N x 2 m = 320 Nm) acting clockwise.
2. The torque due to the tension in the left rope (T1 x 1.2 m = 1.2T1 Nm) acting counterclockwise.
3. The torque due to the tension in the right rope (T2 x 2.8 m = 2.8T2 Nm) acting counterclockwise.
4. The torque due to the weight of the painter (640 N x 0.8 m = 512 Nm) acting clockwise.
Since the platform is in equilibrium, the sum of the torques must be zero. Therefore:
Clockwise torques = Counterclockwise torques.
320 Nm + 512 Nm = 1.2T1 Nm + 2.8T2 Nm.
832 Nm = 1.2T1 Nm + 2.8T2 Nm.
Now, let's consider the vertical forces:
The sum of the forces in the vertical direction must be zero:
T1 + T2 - 160 N - 640 N = 0.
T1 + T2 = 800 N.
We now have a system of equations:
1.2T1 + 2.8T2 = 832,
T1 + T2 = 800.
We can solve this system of equations to find the values of T1 and T2.
Multiplying the second equation by 1.2, we get:
1.2T1 + 1.2T2 = 960.
Subtracting the second equation from the first equation, we get:
1.6T2 = 832 - 960,
1.6T2 = -128,
T2 = -128 / 1.6,
T2 = -80 N.
Substituting this value of T2 into the second equation, we get:
T1 + (-80) = 800,
T1 = 800 + 80,
T1 = 880 N.
Therefore, the tension in the left rope (T1) is 880 N, and the tension in the right rope (T2) is 80 N.
To find the tension in each of the ropes, we need to analyze the forces acting on the system.
Let's consider the forces acting on the platform and the painter. The forces acting on the platform are its weight (160 N) and the tension in the ropes. The forces acting on the painter are his weight (640 N) and the tension in the rope attached to the side he is standing on.
Since the platform is in equilibrium, the sum of the vertical forces acting on it must be zero. This means that the tension in the right rope must be equal to the sum of the weight of the platform and the weight of the painter.
Tension in right rope = weight of platform + weight of painter
Tension in right rope = 160 N + 640 N
Tension in right rope = 800 N
To find the tension in the left rope, we can use the principle of moments. The moment of a force is given by the product of the force and the perpendicular distance from the line of action of the force to the pivot point.
Considering the platform as the pivot point, the clockwise moment due to the tension in the right rope must be balanced by the anticlockwise moment due to the painter's weight.
Clockwise moment = Tension in right rope × Distance between ropes
Anticlockwise moment = Weight of painter × Distance between painter and left rope
Setting these two moments equal, we can solve for the tension in the left rope.
Tension in left rope × Distance between ropes = Weight of painter × Distance between painter and left rope
Tension in left rope × (4 m - 1.2 m) = 640 N × 1.2 m
Tension in left rope × 2.8 m = 768 N·m
Tension in left rope = 768 N·m / 2.8 m
Tension in left rope ≈ 274.29 N
Therefore, the tension in each of the ropes is approximately 800 N in the right rope and 274.29 N in the left rope.