When do you add forces
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A distance of 10.8 cm beyond its natural length will a force of 30 N keep this spring stretched
<u>Explanation:</u>
Work, W = 2 J
Initial distance, = 30 cm
Final distance, = 42 cm
Force, F = 30 N
Stretched length, x = ?
We know,
W = 1/2 kΔx²
Δx = 42-30 cm = 12 cm = 0.12 m
2 J = 1/2 k X (0.12)²
k = 277.77 N/m
According to Hooke's law,
F = kx
30 N = 277.77 X x
x = 0.108 m
x = 10.8 cm
A distance of 10.8 cm beyond its natural length will a force of 30 N keep this spring stretched.
The magnitude of the forces acting at the top are;
= 132.95 N
= 0
The magnitude of the forces acting at the bottom are;
=
= -132.95 N
= 784.8 N
The known parameters in the question are;
The mass of the person, m₁ = 70.0 kg
The length of the ladder, l = 6.00 m
The mass of the ladder, m₂ = 10.0 kg
The distance of the base of the ladder from the house, d = 2.00 m
The point on the roof the ladder rests = A frictionless plastic rain gutter
The location of the center of mass of the ladder, C.M. = 2 m from the bottom of the ladder
The location of the point the person is standing = 3 meters from the bottom
g = The acceleration due to gravity ≈ 9.81 m/s²
The required parameters are;
The magnitudes of the forces on the ladder at the top and bottom
The strategy to be used;
Find the angle of inclination of the ladder, θ
At equilibrium, the sum of the moments about a point is zero
The angle of inclination of the ladder, θ = arccos(2/6) ≈ 70.53 °C
Taking moment about the point of contact of the ladder with the ground, <em>B </em>gives;
= 0
Therefore;
=
Where;
= The sum of clockwise moments about <em>B</em>
= The sum of counterclockwise moments about <em>B</em>
Therefore, we have;
= 2 × (2/6) × 10.0 × 9.81 + 3.0 × (2/6) × 70 × 9.81
=
× √(6² - 2²)
Therefore, we get;
2 × (2/6) × 10.0 × 9.81 + 3.0 × (2/6) × 70 × 9.81 = × √(6² - 2²)
= (2 × (2/6) × 10.0 × 9.81 + 3.0 × (2/6) × 70 × 9.81)/(√(6² - 2²)) ≈ 132.95
The reaction force on the wall, ≈ 132.95 N
We note that the magnitude of the reaction force at the roof, = The magnitude of the frictional force of bottom of the ladder on the floor,
but opposite in direction
Therefore;
=
= -
≈ -132.95 N
Similarly, at equilibrium, we have;
∑Fₓ = = 0
The vertical component of the forces acting on the ladder are, (taking forces acting upward as positive;
= -70.0 × 9.81 - 10 × 9.81 +
∴ The upward force acting at the bottom, = 784.8 N
Therefore;
The magnitudes of the forces at the ladder top and bottom are;
At the top;
=
≈ 132.95 N←
= 0 (The surface upon which the ladder rest at the top is frictionless)
At the bottom;
=
≈ -132.95 N →
=
= 784.8 N ↑
Learn more about equilibrium of forces here;