If the mass of an object is 14 lbm, what is its weight, in lbf, at a location where g = 32.0 ft/s2?
1 answer:
Using the formula:
w = m x g ....... eq1
here w is weight of the object.
m is mass of the object, and
g is the acceleration of gravity.
mass, m = 14 lbm (given)
acceleration of gravity, g= 32.0 ft/
Now, substituting the values in equation (1):
w = 14lbm x 32.0 ft/ = 448 lbm ft/
since, 1 lbf = 32.174 lbm ft/
so, w = 448 x
w = 13.924lbf
Hence, the mass of an object is 13.924 lbf.
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Answer:
The length of the flagpole is approximately 87.43 m
Explanation:
The given parameters of the cable attached to the flagpole are;
The point along the flagpole at which the cable is attached = 32.0 m
The angle with respect to the ground at which the raising of the flagpole is halted = 60.0°
The downward force exerted by the cable, = 1.233 × 10⁴ N
The force exerted by the cable to the left = 1.233 × 10⁴ N
Let 'W' represent the weight of the flagpole, at equilibrium, we have;
The sum of vertical forces = 0
Therefore;
+ W - R = 0
W - R = -1.233 × 10⁴ N
Taking moment about the support at the base of the pole, we get;
1.233 × 10⁴ × d × cos(60.0°) - 1.233 × 10⁴ ×d× sin(60.0°) + W × d/2 ×cos(60.0°) = 0
∴ W × d/2 ×cos(60.0°) ≈ 4513.093·d
W = 2 × 4513.093/(cos(60.0°)) ≈ 18,052.373 N
R = 18,052.373 + 1.233 × 10⁴ ≈ 30,382.373
R ≈ 30,382.373 N
Taking moment about the point of attachment of the cable to the ground, we have;
W × ((d/2) × cos(60.0°) + 32) = R × 32
∴ (d/2) = ((30,382.373 × 32/18,052.373) - 32)/(cos(60.0°)) ≈ 43.71281
d = 2 × 43.71281 ≈ 87.43
The length of the flagpole, d ≈ 87.43 m