1)Three forces act on an object as in the figure. 𝐹⃗1 has a magnitude of 4.00 N and an angle of
60.00. F2 and F3 have magnitudes of 2 N and 5 N, respectively.
a) Determine each force's components:
The angle is 60.00 degrees, and the magnitude is 4.00 N.
𝐹⃗1x = 𝐹1 * cos(θ)
𝐹⃗1x = 4.00 N * cos(60.00°)
𝐹⃗1x = 4.00 N * 0.5
𝐹⃗1x = 2.00 N
𝐹⃗1y = 𝐹1 * sin(θ)
𝐹⃗1y = 4.00 N * sin(60.00°)
𝐹⃗1y = 4.00 N * (√3/2)
𝐹⃗1y = 4.00 N * 1.732 / 2
𝐹⃗1y ≈ 3.464 N
𝐹⃗2x = 2.00 N (no y-component)
𝐹⃗3y = -5.00 N (no x-component)
b) Determine the net force components, magnitude, and angle:
Net 𝐹⃗x = 𝐹⃗1x + 𝐹⃗2x = 2.00 N + 2.00 N = 4.00 N (rightward)
Net 𝐹⃗y = 𝐹⃗1y + 𝐹⃗3y = 3.464 N - 5.00 N = -1.536 N (downward)
Net 𝐹⃗ = √((Net 𝐹⃗x)^2 + (Net 𝐹⃗y)^2)
Net 𝐹⃗ = √((4.00 N)^2 + (-1.536 N)^2)
Net 𝐹⃗ ≈ √(16.00 N^2 + 2.362 N^2)
Net 𝐹⃗ ≈ √18.362 N^2
Net 𝐹⃗ ≈ 4.29 N
θ = atan(Net 𝐹⃗y / Net 𝐹⃗x)
θ = atan(-1.536 N / 4.00 N)
θ ≈ atan(-0.384)
θ ≈ -21.8°
The angle is approximately -21.8 degrees, but it's measured counterclockwise from the positive
x-axis, so the angle with respect to the positive x-axis is 180° - 21.8° = 158.2°.
c) Determine the acceleration components, magnitude, and angle:
𝐹⃗ = m * 𝐚⃗
Acceleration components:
𝐚⃗x = (Net 𝐹⃗x) / m = (4.00 N) / (2 kg) = 2.00 m/s² (rightward)
𝐚⃗y = (Net 𝐹⃗y) / m = (-1.536 N) / (2 kg) ≈ -0.768 m/s² (downward)
𝑎⃗ = √((𝑎⃗x)^2 + (𝑎⃗y)^2)
𝑎⃗ = √((2.00 m/s²)^2 + (-0.768 m/s²)^2)
𝑎⃗ ≈ √(4.00 m²/s⁴ + 0.590 m²/s⁴)
𝑎⃗ ≈ √4.59 m²/s⁴ 𝑎⃗ ≈ 2.14 m/s²
θ = atan(𝑎⃗y / 𝑎⃗x) θ = atan(-0.768 m/s² / 2.00 m/s²) θ ≈ atan(-0.384) θ ≈ -21.8°
The angle of the acceleration vector is approximately -21.8 degrees, which matches the angle of
the net force. So, the acceleration vector makes an angle of 158.2° with the positive x-axis.
2) A person drags a 30.0 kg box across the floor by pulling on a rope at an angle of theta = 20.0
degrees above the horizontal.
a) Sketch the free-body diagram for the system considering the box and the rope as one object:
b) Use Newton's second law to determine unknown forces and accelerations:
Newton's second law states:
𝐹 = m * 𝐚, where 𝐹 is the net force, m is the mass, and 𝐚 is the acceleration.
Weight (𝑊) = m * g 300 N = (30.0 kg) * g g = 10.0 m/s²
Σ𝐹_horizontal = F_person - F_friction - T * cos(θ)
Σ𝐹_horizontal = 160 N - 120 N - T * cos(20.0°)
Σ𝐹_vertical = -W + T * sin(θ)
Σ𝐹_vertical = -300 N + T * sin(20.0°)
For horizontal motion:
Σ𝐹_horizontal = m * 𝐚_horizontal 160 N - 120 N - T * cos(20.0°) = (30.0 kg) * 𝐚_horizontal
For vertical motion:
Σ𝐹_vertical = m * 𝐚_vertical -300 N + T * sin(20.0°) = (30.0 kg) * 𝐚_vertical
c) To find the time it takes for the person to pull the box a distance of 10 m if the box starts from
rest, you need to determine the acceleration and then use kinematic equations.
d = ½ at^2
t = sqrt(2d/ahorizontal)
t=sqrt(2*10m/ahorizontal)
t=sqrt(20m/ahorizontal)
t=sqrt(20m/(30.0kg)*ahorizontal)
t= sqrt(20m/(30.0kg)*(160N-120N-T*cos(20.0 degrees)))
3) F = (5.00 kg) x (9.80 m/s^2) x sin(30 degrees)
Fnet = F - Gravity-parallel
Fnet = 65.0N - (5.00kg) x (9.80 m/s^2) x sin(30.0 degrees)
Fnet = m x a
a = Fnet/m
a = 65 N - 5 kg x 9.8 m/s^2 x sin(30 degrees)/5 kg
a = 65 N - 24.51 N/5 kg
a = 40.49 N/5 kg
a = 8.10 m/s^2