physics torque problem?

Question

A 70 kg painter is painting the wall of a building. He stands on a horizontal board of length 5.0 m and mass 15 kg, suspended from vertical ropes attached to each end. If the painter stands 1.5 m from one end of the board, what would the tensions in the ropes be?

tension in rope nearer to the painter(a)

tension in rope further from the painter(b)

tension in rope nearer to the painter(a)
tension in rope further from the painter(b)

? · Accepted Answer

assume the left rope's tension is T1 and T2 the right one ; painter is 1.5 m away from T1

Momenta balance seen from the left side :
(70*1.5+15*5/2)*9.806 = T2*5
T2 = (70*1.5+15*5/2)*9.806/5 = 279.5 N

Momenta balance seen from the right side :
(70*3.5+15*5/2)*9.806 = T1*5
T1 = (70*3.5+15*5/2)*9.806/5 = 554.0 N

T1+T2 = 554.0+279.5 = 833.5 N 
(mp+mb)*g = (70+15)*9.806 = 833.5 N ...it works !!!

Anonymous · Answer

given
1 = painter
2 = board
L = 5 m
d = 1.5 m
Ta = tension of rope nearer to the painter
Tb = tension of rope further from the painter
must be
Ta + Tb - (m1 + m2) g = 0 [sum of forces = 0]
Tb L -  m1 g d + m2 g L/2 = 0 [sum of momenta = 0]
then
Tb = m1 g d / L + 1/2 m2 g  = 70*9.8*1.5 / 5 + 0.5*15*9.8 = 279.3 N
Ta = (m1 + m2) g - Tb = (70 +15)*9.8 - 279.3 = 553.7 N

Andrew Smith · Answer

You know if a painter did this, and was spotted, worksafe would have his guts for garters.
The plank is unstable, the ropes do not meet the standards for a stress tested support.  There are no handrails.
Inadequate provision to protect passers by underneath.

But to find the answer we need two equations because there are two unknowns.
The first one is F1 + F2 = 85 kg wt  ( together the ropes must support the entire weight)

The second involves torques around any convenient point.
If the board is to NOT rotate then the torques must sum to zero.
ie take the torque around the closer end.  70 * 1.5 + 15 * 2.5 - F2 * 5 = 0
F2 = (70*1.5+15*2.5)/5  = 28.5 kg wt upwards

From equation 1 F1 = 85 - 28.5 = 56.5 kg wt upwards

If you know the value of g then you could convert these to Newtons which is the SI unit of force.

Anonymous · Answer

The weight of the board alone causes a tension of 1/2 (mg) = 1/2 (15 x 9.8) = 73.5N. in each.
(5m/1.5m) = ratio of 3.333:1 painter's weight distribution.
(70 x g)/(3.333 + 1) = 158.32N. tension on the further rope, and (70 x g) - 158.32 = 527.68N. in the nearer rope.
Nearer rope tension = (527.68 + 73.5) = 601.18N.
Further rope tension = (158.32 + 73.5) = 231.82N.

Check, (601.18 + 231.82)/9.8 = 85kg. All good.

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physics torque problem?

4 Answers