A beam pivoted at one end has a force of 5.0N acting vertically upwards on it as shown. the beam is in equilibrium.
what is the weight of the beam.

The weight of the beam at the given equilibrium position is 2.0 N.

The given parameters;

The weight of the beam is calculated by taking moment about the pivot of the beam as follows;

clockwise moment = anticlockwise moment

Thus, the weight of the beam at the given equilibrium position is 2.0 N.

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Option A

Explanation:

Taking moments about the pivot, we know that the sum of clockwise moments equal the sum of anticlockwise moments at equilibrium. Therefore, at the pivot point

5N*2cm=Weight of beam*5 cm

Weight of beam = 5N*2cm/5cm=2 N

Clearly, the weight of the beam is 2N for it to remain in equilibrium

d₁ = displacement of tern from arctic to antarctic = 11000 miles

d₂ = displacement of tern from antarctic to arctic =   - 11000 miles

d = total displacement for the journey

total displacement for the journey is given as

d = d₁ + d₂

d = 11000 + (-11000)

d = 0 miles

t₁ = time of travel from arctic to antarctic = 3 months = 3 x 30 days = 90 days

t₂ = time of travel from antarctic to arctic = 3 months = 3 x 30 days = 90 days

t = total time of travel

total time of travel is given as

t = t₁ + t₂

t = 90 + 90

t = 180 days

average velocity is given as

average velocity = total displacement/ total time

v = d/t

v = 0/180

v = 0 miles/day