A painter is standing on a platform suspended by two ropes. The mass of the pain

Question

A painter is standing on a platform suspended by two ropes. The mass of the painter is 60.0 kg. The mass of the platform (which is uniform) is 25.0 kg. A pail of paint of mass 4.00 kg is located on the platform as shown.

(a) What are the tensions in the two ropes when the painter is located at the right end of the platform?

(b) Show that the system is not in equilibrium when the painter stands at the left end of the platform.

(c) How close to left end can he get while still maintaining equilibrium?

Explanation / Answer

a) balancing moment about left string,

(1 x 4g) + ( 2 x 25g) - (4 x TR) + ( 5 x 60g) = 0

TR = 868.18 N .............tensionin right string

b) if painter goes to left of string then moment due to painter + bucket + platform weight

torque = (1 x 60g ) - (1 x 4g) - (2 x 25g) = 58.86 Nm counterclockwise

and as rope can on only pull not push

so moment due to right about left rope will be counterclockwise.

so it can not be in equilbrium

c) painter to go to maximum distance such that torque calculated above have to be zero .

so
torque = (d x 60g ) - (1 x 4g) - (2 x 25g) = 0

d = 0.9 m form left rope

distance from left end = 0.1 m

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.