A ball is thrown downward with an initial speed of 6m/s. the ball's velocity after 4 seconds is m/s. (g=-9.8m/s^2)
1 answer:
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Answer:
a) the values of the angle α is 45.5°
b) the required magnitude of the vertical force, F is 41 lb
Explanation:
Applying the free equilibrium equation along x-direction
from the diagram
we say
∑Fₓ = 0
Pcosα - 425cos30° = 0
525cosα - 368.06 = 0
cosα = 368.06/525
cosα = 0.701
α = cos⁻¹ (0.701)
α = 45.5°
Also Applying the force equation of motion along y-direction
∑Fₓ = ma
Psinα + F + 425sin30° - 600 = (600/32.2)(1.5)
525sin45.5° + F + 212.5 - 600 = 27.95
374.46 + F + 212.5 - 600 = 27.95
F - 13.04 = 27.95
F = 27.95 + 13.04
F = 40.99 ≈ 41 lb
Answer: 12 N to the right
Explanation:
If we calculate the net force acting on the box, we will have:
<u>In y-component:</u>
(1)
Where is the Normal force, directed upwards and
is the weight of the box (gravity force), directed downwards.
(2)
(3) Hence the net force in the vertical component is zero
<u>In x-component:</u>
(4)
Where and
(5)
(6) This is the net force in the horizontal component
Therefore, the total net force acting on the box is 12 N directed to the right