Asked by Maura
A block with a mass of m = 21 kg rests on a frictionless surface and is subject to two forces acting on it. The first force is directed in the negative x-direction with a magnitude of F1 = 9.5 N. The second has a magnitude of F2 = 19.5 N and acts on the body at an angle θ = 10 ° up from the horizontal as shown.
m = 21 kg
F1 = 9.5 N
F2 = 19.5 N
θ = 10 °
Solve numerically for the block's acceleration in the x-direction, ax, in m/s2.
ax=
m = 21 kg
F1 = 9.5 N
F2 = 19.5 N
θ = 10 °
Solve numerically for the block's acceleration in the x-direction, ax, in m/s2.
ax=
Answers
Answered by
Henry
F = F1 + F2 = -9.5 + 19.5[10o]
F = -9.5 + 19.5*cos10 + (19.5*sin10)
F = -9.5 + 19.20 + 3.39i
F = 9.7 + 3.39i
a = X/m = 9.7/21 = 0.462 m/s^2.
F = -9.5 + 19.5*cos10 + (19.5*sin10)
F = -9.5 + 19.20 + 3.39i
F = 9.7 + 3.39i
a = X/m = 9.7/21 = 0.462 m/s^2.
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