The magnitudes of the forces that the ropes must exert on the knot connecting are :
- F₁ = 118 N
- F₂ = 89.21 N
- F₃ = 57.28 N
<u>Given data :</u>
Mass ( M ) = 12 kg
∅₂ = 63°
∅₃ = 45°
<h3>Determine the magnitudes of the forces exerted by the ropes on the connecting knot</h3><h3 />
a) Force exerted by the first rope = weight of rope
∴ F₁ = mg
= 12 * 9.81 ≈ 118 kg
<u>b) Force exerted by the second rope </u>
applying equilibrium condition of force in the vertical direction
F₂ sin∅₂ + F₃ sin∅₃ - mg = 0 ---- ( 1 )
where: F₃ = ( F₂ cos∅₂ / cos∅₃ ) --- ( 2 ) applying equilibrium condition of force in the horizontal direction
Back to equation ( 1 )
F₂ = [ ( mg / cos∅₂ ) / tan∅₂ + tan∅₃ ]
= [ ( 118 / cos 63° ) / ( tan 63° + tan 45° ) ]
= 89.21 N
<u />
<u>C ) </u><u>Force </u><u>exerted by the</u><u> third rope </u>
Applying equation ( 2 )
F₃ = ( F₂ cos∅₂ / cos∅₃ )
= ( 89.21 * cos 63 / cos 45 )
= 57.28 N
Hence we can conclude that The magnitudes of the forces that the ropes must exert on the knot connecting are :
F₁ = 118 N, F₂ = 89.21 N, F₃ = 57.28 N
Learn more about static equilibrium : brainly.com/question/2952156
Answer:
Acceleration is the rate of change in velocity. Momentum is the mass times the velocity. So if you multiply the mass times the acceleration, you get the of change of momentum.
Definition: Momentum = (mass) x (speed)
OK. Here we go.
Watch closely:
Divide each side
by 'mass' : <span>Momentum / mass = Speed </span>
Did you follow that ?
Answer:
The second dart leaves the gun two times as faster than the first one.
Explanation:
Assuming no energy loss during the spring-dart energy transfer, we have by the conservation of energy principle
Given an arbitrary 
and its double, 
, launch velocities are
