The weight of a box is found to be 30 N. What is the approximate mass of the box?
2 answers:
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Answer:
a) T = (281.47 i ^ + 714.56 j ^) N , b) F_net = (281.47 i ^ - 110.44 j ^) N ,
c) F = 281.70 N, d) θ = 338.58º , e) a = 3,588 m / s² , f) θ = 201.45º
Explanation:
For this exercise we will use Newton's second law on each axis
X axis
-Tₓ = m aₓ
Y Axisy
–W = m a_{y}
Let's use trigonometry to find the components of force
sin 21.5 = Tₓ / T
cos 21.5 = T_{y} / T
Tₓ = T sin 21.5
T_{y} = T cos 21.5
Tₓ = 768 sin 21.5 = 281.47 N
T_{y} = 768 cos 21.5 = 714.56 N
a) the force of the rope on Tarzan is
T = (281.47 i ^ + 714.56 j ^) N
b) The net force is the subtraction of the tension minus the weight of Tarzan
Y Axis F_net = 714.56 - 825 = -110.44 N
F_net = (281.47 i ^ - 110.44 j ^) N
c) Let's use Pythagoras' theorem
F = √ (Fₓ² + T_{y}²)
F = √ (281.47² + 110.44²)
F = 281.70 N
d) Let's use trigonometry
tan θ = F_{y} / Fₓ
θ = tan⁻¹ F_{y} / Fₓ
θ = tan⁻¹ (-110.44 / 281.47)
θ = -21.42º
This angle is average clockwise, for counterclockwise measurement
θ = 360 - 21.42
θ = 338.58º
Acceleration
X axis
Tₓ = m aₓ
aₓ = Tₓ / m
The mass of Tarzan is
m = W / g
m = 825 / 9.8 = 84.18 kg
aₓ = 281.47 / 84.18
aₓ = -3.34 m / s2
Y Axis
T_{y}-W = m a_{y}
a_{y} = (T_{y} -W) / m
a_{y} = (714.56-825) / 84.18
a_{y} = - 1,312 m / s²
Acceleration Module
a = √ aₓ² + a_{y}²
a = √ (3.34² +1.312²)
a = 3,588 m / s²
The angle
θ = tan⁻¹ a_{y} / aₓ
θ = tan⁻¹ (-1312 / -3.34)
θ = 21.45º
Notice that the two components of the acceleration are negative, so the angle is in the third quadrant, to measure from the x-axis
θ = 180 + 21.45
θ = 201.45º