(a) Fx = 1.464 N
(b) Fy = 1.952 N
(c) F(x, y) = 1.464 i + 1.952 j
Given
Mass = 1kg
Acceleration = 2.44 m/s2
Angle with positive X axis = 53°
As we know
F = ma
By substituting value
F= 1×2.44 N
F= 2.44 N
(a) Component of force in X direction
Fx = F Cosθ
Fx = 2.44 Cos(53°)
Fx = 2.44 × 0.60 = 1.464 N
(b) Component of force in Y direction
Fy = F Sinθ
Fy = 2.44 Sin(53°) = 2.44 × 0.80 = 1.952 N
(c) Net force in vector notation
F(x, y) = 1.464 i + 1.952 j
Thus we got net force.
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Answer:
not clear pic...but it's definitely not A)
Answer:
0.54
Explanation:
Draw a free body diagram. There are 5 forces on the desk:
Weight force mg pulling down
Applied force 24 N pushing down
Normal force Fn pushing up
Applied force 130 N pushing right
Friction force Fnμ pushing left
Sum of the forces in the y direction:
∑F = ma
Fn − mg − 24 = 0
Fn = mg + 24
Fn = (22)(9.8) + 24
Fn = 240
Sum of the forces in the x direction:
∑F = ma
130 − Fnμ = 0
Fnμ = 130
μ = 130 / Fn
μ = 130 / 240
μ = 0.54
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