Answer:
angle minimum θ = 41.3º
Explanation:
For this exercise let's use Newton's second law in the condition of static equilibrium
N - W = 0
N = W
The rotational equilibrium condition, where we place the axis of rotation on the wall
We assume that counterclockwise rotations are positive
fr (l sin θ) - N (l cos θ) + W (l/2 cos θ) = 0
the friction force formula is
fr = μ N
fr = μ W
we substitute
μ m g l sin θ - m g l cos θ + mg l /2 cos θ = 0
μ sin θ - cos θ + ½ cos θ= 0
μ sin θ - ½ cos θ = 0
sin θ / cos θ = 1/2 μ
tan θ = 1/2 μ
θ = tan⁻¹ (1 / 2μ)
θ = tan⁻¹ (1 (2 0.57))
θ = 41.3º
Answer
given,
v = 128 ft/s
angle made with horizontal = 30°
now,
horizontal component of velocity
vx = v cos θ = 128 x cos 30° = 110.85 ft/s
vertical component of velocity
vy = v sin θ = 128 x sin 30° = 64 m/s
time taken to strike the ground
using equation of motion
v = u + at
0 =-64 -32 x t
t = 2 s
total time of flight is equal to
T = 2 t = 2 x 2 = 4 s
b) maximum height
using equation of motion
v² = u² + 2 a h
0 = 64² - 2 x 32 x h
64 h = 64²
h = 64 ft
c) range
R = v_x × time of flight
R = 110.85 × 4
R = 443.4 ft
I would make the ramp flatter. In doing so the ramp would have to be longer.
Answer:
b) 4781 N
Explanation:
Because there is a redius do this question is talking about the acceleration force which= mv^2/r
so a=15^2/80=2.8125 m^2/s
so the force will be = m.a
F =1700×2.8125=4781.25 N
Answer:
1. 310 N
2. 310 N
Explanation:
1. Determination of the net force.
Force applied (Fₐ) = 430 N
Force experience (Fₑ) = 120 N
Net force (Fₙ) =?
Fₙ = Fₐ – Fₑ
Fₙ = 430 – 120
Fₙ = 310 N
Hence, the net force acting on the crate is 310 N
2. Determination of the net force.
Force applied (Fₐ) = 375 N
Force of friction (Fբ) = 65 N
Net force (Fₙ) =?
Fₙ = Fₐ – Fբ
Fₙ = 375 – 65
Fₙ = 310 N
Hence, the net force acting on the crate is 310 N.
