Question

A worker stands a distance

d = 0.325 m

from the left end of a plank as shown in the figure. The plank is supported by three cords. Find the tension in each cord (in N). Assume the plank is uniform, with length L = 2.00 m and mass 29.5 kg, and the weight of the worker is 720 N. (Due to the nature of this problem, do not use rounded intermediate values in your calculations—including answers submitted in WebAssign.)

A worker stands on a horizontal 2.00 m long plank a distance d from the left end. The plank is supported by three cords.

• A cord at the right end of the plank goes up and right, making a 40.0° angle with the horizontal and applying a force vector T₁ on the plank outward along the cord.
• A cord at the left end of the plank goes vertically up, applying a force vector T₂ on the plank outward along the cord.
• A cord at the left end of the plank goes horizontally left, applying a force vector T₃ on the plank outward along the cord.

|T₁|

=  
Write the torque equilibrium equation with the rotation axis about the left end of the plank. What are the forces acting in the vertical direction? You will need to write the correct component of T₁.  N

|T₂|

= N

|T₃|

= N

(b)

What If? The worker now begins to walk to the right. If each of the cords can support a maximum tension of 780 N, which cord fails first?

cord 1cord 2     cord 3

At what location does this occur (measured in m from the left end of the plank)?

m

Answer 1

40.0 Tco40 - 2.00 m - W m g T,sin40

Tension force T₁ is resolved into two components, vertical component (T₁ sin40) and

horizontal component ( T₁ cos40 )

At equilibrium, Sum of forces is zero.

Sum of vertical forces is zero, i.e., sum of upward forces equals sum of downward forces.

T₂ + T₁ sin40 = W = 780 N ..................................(1)

Sum of horizontal forces is zero, i.e, sum of forces acting in left equals the sum of forces acting on right.

T₁ cos40 = T₃..........................(2)

At equilibrium, sum of Torques in counterclockwise direction equals sum of Torques in clockwise direction.

Let us consider the torque equations, by considering the rotation axis at left end of plank.

T₁ sin40 $\times$ 2 = w $\times$ d ..................(3)

T₁ sin40 $\times$ 2 = 780 $\times$ 0.325 = 253.5

Hence T₁ = 253.5 / (2 $\times$ sin40 ) = 197.19 N ................(4)

Tension force T₂ is obtained from eqn.(1) using known T₁

T₂ = 780 - T₁ sin40 = 780 - (197.19 sin40 ) = 653.25 N

Tension force T₃ is obtained from eqn.(2), T₃ = T₁ cos40 = 197.19 $\times$ cos40 = 151.06 N

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As the worker walks towards right , distance d increases.

It can be seen from eqn.(3), Tension force T₁ is proportional to d .

Hence as d increases, T₁ increases

It can be seen from eqn. (1), when distance d increases, T₁ increases and that makes T₂ to decrease.

From eqn.(2), we have T₁ > T₃ always.

Due to all these reasons, T₁ is the tension force that is going to exceed the limit 780 N

The distance d for which the right side cord breaks is obtained from eqn.(3), by substituting T₁ = 780

d = 2 T₁ sin40 / w = (2 $\times$ 780 $\times$ sin40)/780 = 1.286 m