Let mAngleA = 40°. If AngleB is a complement of AngleA, and AngleC is a supplement of AngleB, find these measures.
2 answers:
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Answer:
1) Please find attached, created with Microsoft Visio
2) The acceleration of the masses connected by the light string is 0.00735 m/s²
3) The tension in the cord is 0.147 N
4) The time it would take the block to go 1.2 m to the edge of the table is approximately 18.07 s
5) The velocity of the cart as soon as it gets to the edge of the table is 0.042 m/s
Explanation:
1) Please find attached, the required free body diagram, showing the tension, weight and frictional (zero friction) forces acting on the cart and the mass created with Microsoft Visio
2) The acceleration of the masses connected by the light string is given as follows;
F = Mass, m × Acceleration, a
The mass of the truck, M = 20.0 kg
The mass attached to the string, hanging rom the pulley, m = 0.0150 kg
The force, F acting on the system = The pulling force on the cart = The tension on the cable = The weight of the hanging mass = 0.0150 × 9.8 = 0.147 N
The pulling force acting on the cart, F = M × a
∴ F = 0.147 N = 20.0 kg × a
a = 0.147 N/(20.0 kg) = 0.00735 m/s²
The acceleration of the truck = a = 0.00735 m/s²
3) The tension in the cord = F = 0.147 N
4) The time, t, it would take the block to go 1.2 m to the edge of the table is given by the kinematic equation, s = u·t + 1/2·a·t²
Where;
s = The distance to the edge of the table = 1.2 m
u = The initial velocity = 0 m/s (The cart is assumed to be initially at rest)
a = The acceleration of the cart = 0.00735 m/s²
t = The time taken
Substituting the known values, gives;
s = u·t + 1/2·a·t²
1.2 = 0 × t + 1/2 ×0.00735 × t²
1.2 = 1/2 ×0.00735 × t²
t² = 1.2/(1/2 ×0.00735) ≈ 326.5306
t = √(1.2/(1/2 ×0.00735)) ≈ 18.07
The time it would take the block to go 1.2 m to the edge of the table = t ≈ 18.07 s
5) The velocity, v, of the cart as soon as it gets to the edge of the table is given by the kinematic equation, v² = u² + 2·a·s as follows;
v² = u² + 2·a·s
u = 0 m/s
v² = 0² + 2 × 0.00735 × 1.2 = 0.001764
v = √(0.001764) = 0.042
The velocity of the cart as soon as it gets to the edge of the table = v = 0.042 m/s.
To solve the problem, it is necessary to use the concepts of gravitational force, centripetal force and trigonometric components that can be extrapolated from the statement.
By definition we know that the Force of Gravity is given by
Where,
m= Mass
g = Gravitational Acceleration
The centripetal force is given by,
Where,
m = Mass
v = Velocity
R = Radius
For the case described in the problem, the Force of gravity the net component would be given by sin?, While for the centripetal force the net component is in the horizontal direction, therefore it corresponds to the
Then,
From the radius we have its length but not the net height, which would be given by
So equating the equations we have to
Re-arrange to find v,
Replacing with our values
Therefore the tangential velocity of the mass is 0.624m/s