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2 answers:
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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:
(i) See attached image for the drawing
(ii) net force given in component form: (20, 20)N with magnitude:
Explanation:
First try to write all forces in vector component form:
The force F1 acting at 45 degrees would have multiplication factors of on both axes, to take care of the sine and cosine projections. Therefore, the:
x-component of F1 is
y-component of F1 is
As far as force F2, it is given already in x and y components, then:
x-component of F2 = 8 N
y-component of F2 = -6 N (negative meaning pointing down the y-axis)
Force F3 has only component (upwards) in the y-direction
x-component of F3 = 0 N
y-component of F3 =14 N
The additions of all these component by component, gives the resultant force (R) acting on the 5 kg mass:
x-component of R = 12 + 8 = 20 N
y-component of R = 12 + 14 - 6 = 20 N
Therefore, the acceleration that the mass receives due to this force is given in component form as:
x-component of acceleration: 20 N / 5 kg =
y-component of acceleration: 20 N / 5 kg =
Now we can calculate the components of the velocity of this mass after 2 seconds of being accelerated by this force, using the formula of acceleration times time:
x-component of the velocity is:
y-component of the velocity is:
