A woman holds a 1.9 m long uniform 5.5 kg pole as shown in the figure .
her hands are 30 cm apart.
Determine the magnitudes of the forces she must exert with each hand.
To what position should she move her left hand so that neither hand has to exert a force greater than 110 N?
To what position should she move her left hand so that neither hand has to exert a force greater than 35 N?
as shown in the figure...?
I don't think you really need the figure.It's just a womand holding a pole with her hand 30 cm apart. I don't know how to get the figure on here.
You have to have position to determine the sum of moments.
o ok... she is holding it horizontally with her right hand at the end of the pole and her left hand 30 cm away.
To determine the magnitudes of the forces the woman must exert with each hand, we need to analyze the torque equation. Torque is the product of force and the perpendicular distance from the pivot point to the line of action of the force.
Let's denote the distance from the woman's left hand to the pivot point as d₁ and the distance from her right hand to the pivot point as d₂. The total torque acting on the pole must be zero, as it is in equilibrium.
1. Determine the magnitudes of the forces she must exert with each hand:
To find the magnitudes of the forces, we need to set up an equation based on the condition of equilibrium for torques:
T₁ = T₂
Here, T₁ and T₂ represent the torques exerted by her left and right hand, respectively.
The torque exerted by a force is given by:
Torque = Force × Distance × sin(θ)
Since the forces are perpendicular to the pole, θ is 90 degrees, and sin(θ) = 1.
Thus, we can rewrite the equation as:
F₁ × d₁ = F₂ × d₂
Now let's substitute the given values into the equation:
F₁ × (0.3 m) = F₂ × (1.6 m)
Since we want to determine the magnitudes of the forces, we only need the ratios of the forces. Thus, we can rewrite the equation as:
F₁/F₂ = (1.6 m)/(0.3 m) = 5.3
Now we have a ratio between the magnitudes of the forces. We can choose any value for one of the forces and calculate the other. Let's assume F₂ = 1 N:
F₁/F₂ = 5.3
F₁/1 = 5.3
Therefore, the magnitude of the force exerted by her right hand (F₂) is 1 N, and the magnitude of the force exerted by her left hand (F₁) is 5.3 times greater, which is approximately 5.3 N.
2. To determine the position where neither hand has to exert a force greater than 110 N:
Since the position of the left hand (d₁) is mentioned, we can substitute the given values into the equation from step 1:
(110 N) × d₁ = 5.3 N × (1.6 m)
Now, rearrange the equation to calculate the distance d₁:
d₁ = (5.3 N × 1.6 m) / (110 N)
Simplifying the equation, we get:
d₁ ≈ 0.077 m or 7.7 cm
Therefore, the woman should move her left hand approximately 7.7 cm away from the pivot point so that neither hand has to exert a force greater than 110 N.
3. To determine the position where neither hand has to exert a force greater than 35 N:
Using a similar approach as in step 2, substitute the given values into the equation from step 1:
(35 N) × d₁ = 5.3 N × (1.6 m)
Rearrange the equation to calculate the distance d₁:
d₁ = (5.3 N × 1.6 m) / (35 N)
Simplifying the equation, we get:
d₁ ≈ 0.243 m or 24.3 cm
Therefore, the woman should move her left hand approximately 24.3 cm away from the pivot point so that neither hand has to exert a force greater than 35 N.