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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The speed of the brick dropped by the builder as it hits the ground is 17.32m/s.
Given the data in the question;
Since the brick was initially at rest before it was dropped,
- Initial Velocity;
- Height from which it has dropped;
- Gravitational field strength;
Final speed of brick as it hits the ground;
velocity is simply the same as the speed at which a particle or object moves. It is the rate of change of position of an object or particle with respect to time. As expressed in the Third Equation of Motion:
Where v is final velocity, u is initial velocity, h is its height or distance from ground and g is gravitational field strength.
To determine the speed of the brick as it hits the ground, we substitute our giving values into the expression above.
Therefore, the speed of the brick dropped by the builder as it hits the ground is 17.32m/s.
Learn more about equations of motion: brainly.com/question/18486505