Question

A worker stands a distance

d = 0.250 m

from the left end of a beam as shown in the figure. The beam is supported by three ropes. Find the tension in each rope (in N). Assume the beam is uniform, with length L = 2.00 m and mass 34.0 kg, and the weight of the worker is 710 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 beam a distance d from the left end. The beam is supported by three ropes.

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

|T₁|

= N

|T₂|

= N

|T₃|

= N

(b)

What If? The worker now begins to walk to the right. If each of the ropes can support a maximum tension of 820 N, which rope fails first?

rope 1rope 2    rope 3

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

m

Answer 1

a)
let
d = 0.250 m
M = 34.0 kg
L = 2.00 m
w = 710 N

As the beam is in equilibrium net force and net torque actin on it must be zero.

Apply, net torque about left = 0

T1*L*sin(40) - M*g*(L/2)*sin(90) - w*d*sin(90) = 0

T1 = (M*g*(L/2) + W*d )/(L*sin(40))

= (34*9.8*(2/2) + 710*0.25 )/(2*sin(40))

= 397 N <<<<<<<<-------------------Answer

Now apply, Fnety = 0

T2 + T1*sin(40) - M*g - w = 0

T2 = M*g + w - T1*sin(40)

= 34*9.8 + 710 - 397*sin(40)

= 788 N <<<<<<<<-------------------Answer

Now apply, Fnetx = 0

T1*cos(40) - T3 = 0

T3 = T1*cos(40)

= 397*cos(40)

= 304 N <<<<<<<-------------------Answer

b) rope 1

let x is the distance from left end of the beam to the person when the rope 1 breaks.

now apply net torque about left end = 0

T1*L*sin(40) - M*g*(L/2)*sin(90) - w*x*sin(90) = 0

820*2*sin(40) - 34*9.8*1 - 710*x = 0

==> x = 1.01 m   <<<<<<<<-------------------Answer