Three persons wants to push a wheel cart in the direction marked x in Fig. 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 = 109.5 N ,     θ₃ = 38.35,  b)   m = 0.429 kg

Explanation:

For this force addition exercise, let's decompose the forces into a coordinate system

strength 1

          cos 70 = F₁ₓ / F₁

          sin 70 = / F₁

          F₁ₓ = F₁ cos 70

         F_{1y} = F₁ sin 70

          F₁ₓ = 45 cos 70 = 15.39 N

          F_{1y} = 45 sin70 = 42.29 N

Strength 2

         cos 20 = F₂ₓ / F₂

         sin 20 = F_{2y} / F₂

         F₂ₓ = F₂ cos 20

         F_{2y} = F₂ sin20

         F₂ₓ = 75 cos 20 = 70.48 N

         F_{2y} = 75 sin 20 = 25.65 N

a) in the first part we are asked to find the magnitude of force 3 so that the body is in equilibrium

X axis

         ∑ Fₓ = 0

          F₁ₓ + F₂ₓ + F₃ₓ = 0

          F₃ₓ = - F₁ₓ - F₂ₓ

          F₃ₓ = - 15.39 - 70.48

          F₃ₓ = - 85.87 N

Y Axis

         ∑ F_{y}  = 0

         F_{1y} + F_{2y} +F_{3y}  = 0

         F_{3y} = - F_{1y} -F_{2y}

         F_{3y} = - 42.29 - 25.65

         F_{3y} = - 67.94 N

the magnitude of this force can be found using the Pythagorean theorem

         F₃ = √ (F₃ₓ² + F_{3y}^2)

         F₃ = √ (85.87² + 67.94²)

         F₃ = 109.5 N

The direction of this force can be found using trigonometry

          tan θ =  F₃ = / F₃ₓ

          θ₃ = Tan⁻¹ (67.94 / (85.87))

          θ₃ = 38.35º

since the two vectors are negative this angle is in the third quadrant, measured from the positive side of the x axis is

          θ₃ = 180 + 38.53

          θ₃ = 218.35º

b) I think you have an error in your statement, we have two possibilities:

1) If we eliminate the third force, what is the mass

it should say: "if we apply the third force force ...", they ask that we find the masses of the vehicle, let's use Newton's second law

X axis

       F₁ₓ + F₂ₓ = m a

       15.39 + 70.48 = m 200

        m = 85.87 / 200

        m = 0.429 kg

2) If the force F3 is in the first quadrant

        F1x + F2x + F3x = m a

        m = (15.39 + 70.48 + 85.87) / 200

        m = 0.8587 kg

c

explanation:

i believe the answer is c and this is only worth 10 points you might wanna change it