A person carries a plank of wood 2 m long with one hand pushing down on it at one end with a force F1 and the other hand holding it up at 50 cm from the end of the plank with force F2. If the plank has a mass of 20 kg and its center of gravity is at the middle of the plank, what are the forces F1 and F2?

To begin, let's calculate the position of the center of gravity (CG) of the plank. Since the plank is 2 m long and the CG is at the middle, the CG would be at a distance of 1 m from either end.

Next, let's establish the equation of torque based on the principle of rotational equilibrium. Torque is equal to the force multiplied by the perpendicular distance from the pivot point. The pivot point in this case is the center of gravity of the plank.

For F1, we have:

Torque1 = F1 * d1

For F2, we have:

Torque2 = F2 * d2

Where d1 is the distance from the pivot point (CG) to the hand applying force F1 and d2 is the distance from the pivot point (CG) to the hand applying force F2.

Since F1 is pushing down on one end of the plank, the distance from the CG to the hand applying F1 is 1 m. Therefore, d1 = 1 m.

Since F2 is holding up the plank 50 cm from the end, the distance from the CG to the hand applying F2 is 1 m + 0.5 m = 1.5 m. Therefore, d2 = 1.5 m.

In rotational equilibrium, the sum of the torques acting on an object is equal to zero. Therefore, we can write:

Torque1 + Torque2 = 0

F1 * d1 + F2 * d2 = 0

Substituting the values of d1 = 1 m and d2 = 1.5 m:

F1 * 1 m + F2 * 1.5 m = 0

F1 + 1.5F2 = 0

We also know that the weight of the plank (mg) is equal to the sum of the forces F1 and F2. The weight of the plank can be calculated using the formula:

Weight = mass * acceleration due to gravity

Therefore:

Weight = 20 kg * 9.8 m/s^2 = 196 N

Since F1 and F2 are the respective forces pushing down and holding up the plank, we can write:

F1 + F2 = Weight

F1 + F2 = 196 N

Now, we have a system of two equations:

F1 + 1.5F2 = 0 (Equation 1)
F1 + F2 = 196 N (Equation 2)

To solve this system, we can use elimination or substitution.

Let's solve using substitution:

From Equation 2:
F1 = 196 N - F2

Substituting this value into Equation 1:

(196 N - F2) + 1.5F2 = 0

196 N - F2 + 1.5F2 = 0

0.5F2 = 196 N

F2 = 196 N / 0.5

F2 = 392 N

Now substituting F2 into Equation 2 to find F1:

F1 + F2 = 196 N

F1 + 392 N = 196 N

F1 = 196 N - 392 N

F1 = -196 N

Therefore, the force F1 pushing down on one end of the plank is -196 N, and the force F2 holding up the plank is 392 N.

Please note that the negative sign for F1 indicates that it is acting in the opposite direction. In other words, F1 is pushing down on the plank while F2 is holding it up.

To find the forces F1 and F2, we can start by analyzing the rotational equilibrium of the plank. Since the plank is not rotating, the sum of the torques acting on it must be zero.

The torque τ about a point is given by the formula τ = r * F * sin(θ), where r is the perpendicular distance from the point to the line of action of the force, F is the magnitude of the force, and θ is the angle between the force direction and the line connecting the point and the force.

Let's assign the origin of our coordinate system to be at the point where the person is pushing the plank with force F1. The length of the plank is 2 m, so the distance from the push point to the center of gravity is 1 m. The distance from the center of gravity to the point where the second hand is holding the plank is 50 cm, which is equivalent to 0.5 m. We can now set up our equation based on the torque balance:

τ1 + τ2 = 0

Since the forces F1 and F2 are acting in opposite directions, their torques will have opposite signs. Therefore, we can write:

-F1 * 1 m + F2 * 0.5 m = 0

Simplifying the equation, we find:

-F1 + 0.5F2 = 0

Now, to find the relationship between F1 and F2, we need to consider the vertical equilibrium of the plank. The sum of the forces in the vertical direction must be zero since the plank is not accelerating vertically.

Taking upward as the positive direction, we can set up our equation:

F1 - F2 - mg = 0

Here, m is the mass of the plank, which is given as 20 kg, and g is the acceleration due to gravity (9.8 m/s²).

Substituting the equation F1 - F2 = mg, we have:

mg - F2 = 0.

Now we have a system of two equations:

-F1 + 0.5F2 = 0 (equation 1)
mg - F2 = 0 (equation 2)

To solve this system of equations, we can first solve equation 2 for F2:

F2 = mg

Substituting this into equation 1, we get:

-F1 + 0.5mg = 0

Solving for F1, we have:

F1 = 0.5mg

Plugging in the values for mass m and acceleration due to gravity g, we can calculate the forces:

F1 = 0.5 * 20 kg * 9.8 m/s² = 98 N
F2 = 20 kg * 9.8 m/s² = 196 N

Therefore, the force F1 is 98 N and the force F2 is 196 N.

F₁+mg=F₂ ….(1)

2•F₁+1•mg=1.5•F₂…(2)
From (1)
F₁ =F₂ -mg ….(3)
Substitute (3) in (2)
2(F₂ -mg) +mg = 1.5•F₂
F₂=2mg=....
F₁=mg= ....

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