A steel tank of weight 600 lb is to be accelerated straight upward at a rate of 1.5 ft/sec2. Knowing the magnitude of the force
P is 525 lb, determine the values of the angle a and the required magnitude of the vertical force, F.
1 answer:
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
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Answer:
they meet from the girl's original position at: 2.37 (meters)
Explanation:
We need to use the Newton's law, exactly the second law that relate force, mass and acceleration as: with this we can get both accelerations; solving for acceleration
. Now
and
. Then knowing that they both travel at the same time and assuming that the distance among the girl and the sled is: 15.0-x, so,
and
, solving for the time we get:
and
with this equations we solving for the x that is the distance between the girl and the sled after the apply the force and we get:
. Finally we get:
and replacing the values we have got:
so
so x=2.37 (meters).