Answer:
303.29N and 1.44m/s^2
Explanation:
Make sure to label each vector with none, mg, fk, a, FN or T
Given
Mass m = 68.0 kg
Angle θ = 15.0°
g = 9.8m/s^2
Coefficient of static friction μs = 0.50
Coefficient of kinetic friction μk =0.35
Solution
Vertically
N = mg - Fsinθ
Horizontally
Fs = F cos θ
μsN = Fcos θ
μs( mg- Fsinθ) = Fcos θ
μsmg - μsFsinθ = Fcos θ
μsmg = Fcos θ + μsFsinθ
F = μsmg/ cos θ + μs sinθ
F = 0.5×68×9.8/cos 15×0.5×sin15
F = 332.2/0.9659+0.5×0.2588
F =332.2/1.0953
F = 303.29N
Fnet = F - Fk
ma = F - μkN
a = F - μk( mg - Fsinθ)
a = 303.29 - 0.35(68.0 * 9.8- 303.29*sin15)/68.0
303.29-0.35( 666.4 - 303.29*0.2588)/68.0
303.29-0.35(666.4-78.491)/68.0
303.29-0.35(587.90)/68.0
(303.29-205.45)/68.0
97.83/68.0
a = 1.438m/s^2
a = 1.44m/s^2
Answer:
α = 141.5° (counterclockwise)
Explanation:
If
q₁ = +q
q₂ = -q
q₃ < 0
b = 2*a
We apply Coulomb's Law as follows
F₁₃ = K*q₁*q₃ / d₁₃² = + K*q*q₃ / (2*a)² = + K*q*q₃ / (4*a²)
F₂₃ = K*q₂*q₃ / d₂₃² = - K*q*q₃ / (5*a²)
(d₂₃² = a² + (2a)² = 5*a²)
Then
∅ = tan⁻¹(2a/a) = tan⁻¹(2) = 63.435°
we apply
F₃x = - F₂₃*Cos ∅ = - (K*q*q₃ / (5*a²))* Cos 63.435°
⇒ F₃x = - 0.0894*K*q*q₃ / a²
F₃y = - F₂₃*Sin ∅ + F₁₃
⇒ F₃y = - (K*q*q₃ / (5*a²))* Sin 63.435° + (K*q*q₃ / (4*a²))
⇒ F₃y = 0.0711*K*q*q₃ / a²
Now, we use the formula
α = tan⁻¹(F₃y / F₃x)
⇒ α = tan⁻¹((0.0711*K*q*q₃ / a²) / (- 0.0894*K*q*q₃ / a²)) = - 38.5°
The real angle is
α = 180° - 38.5° = 141.5° (counterclockwise)
You will get 20460000 as your answer which is broken down into, 2.046 x 10^7 as your number has to be between 1-10.
Answer: reaction force = -558N
Explanation:
w = f = 558N
since action force and reaction force are equal in magnitude and opposite in direction,
reaction force = -(f)
reaction force = -558N
if that helps.
