Three persons wants to push a wheel cart in the direction marked x in Fig. [04 Marks] The two person push with horizontal forces F1 and F2 as F1 = 45 N 70degree And F2 = 75 N , 20 Degree (a) Find the magnitude and direction of the force that third person should exert to stop this cart. You can ignore the effects of friction. (b) If the third person exerts the force found in part (a), the cart accelerates at 200 m/S2 in the (+) x-direction. What is the weight of the cart?

a) F₃ = 108.71 N ,    θ = 218.68º    b)  m = 0.8487 kg

Explanation:

a) To solve this problem we will use Newton's second law, in the first part the accuracy is zero

          ∑ F = 0

X axis

          F₁ₓ + F₂ₓ + F₃ₓ = 0

          F₃ₓ = - F₁ₓ -F₂ₓ

Y axis

          + F_{2y} + F_{3y} = 0

         F3y = -F_{1y} -F_{2y}

Let's use trigonometry to find the components of the applied forces

         cos 70 = F₁ₓ / F₁                 F₁ₓ = F₁ cos 70

         sin 70 = F_{1y} / F₁             F_{1y} = F₁ sin 70

         cos 20 = F₂ₓ / F2               F₂ₓ = F₂ cos 70

         sin 20 = F_{2y} / F₂           F_{2y} = F₂ sin 70

         F₁ₓ = 45 cos 70 = 15.39 N

         F_{1y} = 45 sin 70 = 42.29 N

         F₂ₓ = 75 cos 20 = 70.48 N

         F_{2y} = 75 sin 20 = 25.65 N

let's calculate

        F₃ₓ = -15.39 - 70.48

        F₃ₓ = -85.87 N

        F_{3y} = -42.29 - 25.65

        F_{3y} = -67.94 N

this is the force that must be applied to keep the system in equilibrium, we can give this value in the form of module and angle

         F₃ = √ (F₃ₓ² + F_{3y}²)

         F₃ = √ (85.87²  +67.94²)

         F₃ = 108.71 N

         tan θ = F₃ = / F₃ₓ

         θ = tan⁻¹ (67.94 / 84.87)

         θ = 38.68º

Since the two values are negative, this angle is in the third quadrant, therefore the measuring angle n counterclockwise from the positive side of the x-axis is

        θ = 180 + 38.68

        θ = 218.68º

b) It is not clear, but we must suppose that the person exerts this force F3 in the direction of movement, bone at an angle θ= 39.68º, in this case Newton's equation for the x axis remains

         F₁ₓ + F₂ₓ + F₃ₓ = m a

         m = (F₁ₓ + F₂ₓ + F₃ₓ) / a

let's calculate

          m = (15.39 + 7048 + 85.87) / 200

          m = 0.8487 kg

47r7eudueueueurudueu37373