Asked by Anonymous
There are two forces on the 2.36 kg box in the overhead view of Fig. 5-31 but only one is shown. For F1 = 18.6 N, a = 11.4 m/s2, and θ = 26.7°, find the second force (a) in unit-vector notation and as (b) a magnitude and (c) a direction. (State the direction as a negative angle measured from the +x direction.)
Answers
Answered by
Yi
a)
Fnet = ma
F1 + F2 = ma
18.6N + F2 = 2.36(11.4*cos26.7° + 11.4*sin26.7°)
F2 = 42.6i + 12.1j
b)
sqrt(42.6^2 +12.1^2)
= 44.3 (N)
c)
θ = 26.7° + 90°
θ = -116.7°
Fnet = ma
F1 + F2 = ma
18.6N + F2 = 2.36(11.4*cos26.7° + 11.4*sin26.7°)
F2 = 42.6i + 12.1j
b)
sqrt(42.6^2 +12.1^2)
= 44.3 (N)
c)
θ = 26.7° + 90°
θ = -116.7°
Answered by
Yi
a)
Fnet = ma
F1 + F2 = ma
18.6N + F2 = 2.36(11.4*cos26.7° + 11.4*sin26.7°)
F2 = 42.6i + 12.1j
b)
sqrt(42.6^2 +12.1^2)
= 44.3 (N)
c)
θ = 26.7° + 180°
θ = 206.7°
206.7° - 300°
θ = -93.3°
Fnet = ma
F1 + F2 = ma
18.6N + F2 = 2.36(11.4*cos26.7° + 11.4*sin26.7°)
F2 = 42.6i + 12.1j
b)
sqrt(42.6^2 +12.1^2)
= 44.3 (N)
c)
θ = 26.7° + 180°
θ = 206.7°
206.7° - 300°
θ = -93.3°
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