Answer:
Explanation:
Given
Initial reading on scale =40 N
So, we can conclude that weight of the sack is 40 N
After this a 10 N force is applied upward on the sack such that the net force becomes (40-10) N downward (because downward force is more)
This net downward force is the resultant of earth graviational pull and the applied upward force.
So, this downward force acts on the machine which inturn applies an upaward force of same magnitude called Normal reaction.
This situation can be diagramatically represented by figure given below
Answer:
upward force acting = 261.6 N
Explanation:
given,
mass of gibbon = 9.4 kg
arm length = 0.6 m
speed of the swing
net force must provide
force of gravity = - mg
=
=
=9 x 29.067
= 261.6 N
upward force acting = 261.6 N
Answer:
w = 0.189 rad/ s
Explanation:
This exercise we work with the conservation of the moment, the system is made up of the merry-go-round and the child, for which we write the moment of two instants
Initial
L₀ = I₀ w₀
Final
= I w
L₀ = L_{f}
I₀ w₀ = I_{f} w
.w = I₀/I_{f} w₀
The initial moment of inertia is
I₀ = 500 kg. m2
The final moment of inertia
= 500 + m r²
I_{f} = 500 + 20 1.5
I_{f} = 530 kg m²
Initial angular velocity
w₀ = 0.20 rad / s
Let's calculate
w = 500/530 0.20
w = 0.189 rad / s