Two people are carrying a uniform 704.0 N log through the forest. Bubba is 2.2 m from one end of the log (x), and his partner is 0.9 m from the other end (y). The log is 6.2 m long (z). What weight is Bubba supporting

Answer:

F₁ = 499.61 N , this is the force that Bubba support

Explanation:

The trunk is in equilibrium with the two forces applied by man, let’s use the equilibrium relation

let’s set a reference frame at the extreme left and assume that the counterclockwise rotations are positive

Let’s write the expression for the translational equilibrium

subscript 1 is for Bubba’s mass and subscript 2 for his partner

F₁ + F₂ -W = 0

F₁ + F₂ = W

the expression for rotational equilibrium

∑ τ = 0

F₁ 2.2 + F₂ (6.2-0.9) – W 6.2/2 = 0

2.2 F1 + 5.3 F2 = 3.1 W

let’s write our system of equations

F₁ + F₂ = W

2.2 F₁ + 5.3 F₂ = 3.1 W

we solve for F₁ in the first equation and substitute in the second

Answer:F₁ = 499.61 N , this is the force that Bubba support

Explanation:The trunk is in equilibrium with the two forces applied by man, let’s use the equilibrium relation

let’s set a reference frame at the extreme left and assume that the counterclockwise rotations are positive

Let’s write the expression for the translational equilibrium

subscript 1 is for Bubba’s mass and subscript 2 for his partner

F₁ + F₂ -W = 0

F₁ + F₂ = W

the expression for rotational equilibrium

∑ τ = 0

F₁ 2.2 + F₂ (6.2-0.9) – W 6.2/2 = 0

2.2 F1 + 5.3 F2 = 3.1 W

let’s write our system of equations

F₁ + F₂ = W

2.2 F₁ + 5.3 F₂ = 3.1 W

we solve for F₁ in the first equation and substitute in the second

F₁ = W-F₂

2.2 (W- F₂) + 5.3 F₂ = 3.1 W

F₂ ( -2.2 +5.3) = W (3.1 – 2.2)

F₂ = 704 0.9 / 3.1

F₂ = 204.39 N

This is the force that the partner supports

we look for F1

F₁ = W-F₂

F₁ = 704 – 204.39

F₁ = 499.61 N

This is the force that Bubba support