A 50-kg person stands 1.5 m away from one end of a uniform 6.0-m-long scaffold of mass 70.0 kg.
1 answer:
Answer
given,
mass of the person, m = 50 Kg
length of scaffold = 6 m
mass of scaffold, M= 70 Kg
distance of person standing from one end = 1.5 m
Tension in the vertical rope = ?
now equating all the vertical forces acting in the system.
T₁ + T₂ = m g + M g
T₁ + T₂ = 50 x 9.8 + 70 x 9.8
T₁ + T₂ = 1176...........(1)
system is equilibrium so, the moment along the system will also be zero.
taking moment about rope with tension T₂.
now,
T₁ x 6 - mg x (6-1.5) - M g x 3 = 0
'3 m' is used because the weight of the scaffold pass through center of gravity.
6 T₁ = 50 x 9.8 x 4.5 + 70 x 9.8 x 3
6 T₁ = 4263
T₁ = 710.5 N
from equation (1)
T₂ = 1176 - 710.5
T₂ = 465.5 N
hence, T₁ = 710.5 N and T₂ = 465.5 N
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Answer:
a) correct answer is C , b) 14º from the west to the north, c) v_{1g} = 300.79 km / h
Explanation:
This is a relative speed exercise using the addition of speeds.
1) when it is not specified regarding what is being measured, the medicine is carried out with respect to the Z Earth, therefore the correct answer is C
2 and 3) In this case we must compose the speed using the Pythagorean Theorem.
² =
² +
²
where v_{1a} is the speed of the airplane with respect to the air, v_{1g} airplane speed with respect to the Earth, v_{ag} air speed with respect to the Earth
in this case let's clear the speed of the airplane with respect to the Earth
v_{1g} = √(v_{1a}² - v_{ag}²)
v_{1g} = √ (310² - 75²)
v_{1g} = 300.79 km / h
we find the direction of the airplane using trigonometry
sin θ = v_{ag} / v_{1a}
θ = sin⁻¹ (v_{ag} /v_{1a})
θ = sin⁻¹ (75/310)
θ= 14º
the pilot must direct the aircraft at an angle of 14º from the west to the north