A uniform plank of length 4.1 m and weight 240 N rests horizontally on two supports, with 1.1 m of the plank hanging over the right support. To what distance x can a person who weighs 445 N walk on the overhanging part of the plank before it just begins to
tip?
(3.1/4.1)*240*(3.1/2) = (1/4.1)*240*(1.1/2) + 445x
x = 0.56
To determine the distance x a person can walk on the overhanging part of the plank before it begins to tip, we need to calculate the point at which the plank starts to tip.
First, let's find the center of mass of the plank. Since the plank is uniform, the center of mass is located in the middle. Therefore, the center of mass is at a distance of L/2 = 4.1 m / 2 = 2.05 m from the left support.
Now, we need to compare the torques on each side of the fulcrum (the left support) to determine when the plank begins to tip.
The torque on the left side is given by:
Torque_left = weight_of_plank * distance_from_fulcrum = (240 N) * (2.05 m)
The torque on the right side is given by:
Torque_right = weight_of_person * distance_from_fulcrum = (445 N) * (4.1 m - x)
For the plank to remain in equilibrium (not tipping), the torques on both sides of the fulcrum must be equal.
Setting Torque_left = Torque_right, we get:
(240 N) * (2.05 m) = (445 N) * (4.1 m - x)
Simplifying the equation, we have:
(240 N) * (2.05 m) = (445 N) * (4.1 m) - (445 N) * (x)
Solving for x, we can rearrange the equation to isolate x:
(445 N) * x = (445 N) * (4.1 m) - (240 N) * (2.05 m)
Dividing both sides of the equation by (445 N), we get:
x = [(445 N) * (4.1 m) - (240 N) * (2.05 m)] / (445 N)
Calculating this expression, we find:
x = (1824.5 N·m - 492 N·m) / 445 N
x ≈ 2.65 m
Therefore, a person who weighs 445 N can walk approximately 2.65 m on the overhanging part of the plank before it begins to tip.
To determine the maximum distance a person can walk on the overhanging part of the plank before it starts to tip over, we need to consider the torque and balance of forces.
1. First, let's calculate the weight of the plank.
Weight = Mass × Acceleration due to gravity
Given that the weight of the plank is 240 N, we need to find the mass of the plank.
240 N = Mass × 9.8 m/s^2
Mass = 240 N / 9.8 m/s^2 = 24.49 kg
2. Next, let's consider the forces acting on the plank.
a) The weight of the plank (240 N) acts downward at its center of mass, which is located at the midpoint of the plank (2.05 m from either support).
b) The reaction forces from the supports act upward to counterbalance the weight of the plank. Let's denote the left support as "A" and the right support as "B". The reaction force at support A counters half of the plank's weight, while the reaction force at support B counters the other half.
Reaction force at support A = 240 N / 2 = 120 N
Reaction force at support B = 240 N / 2 = 120 N
3. Now, let's determine the maximum torque the person can apply without tipping the plank. The torque acts in the clockwise direction (as viewed from the right support) and is generated by the person's weight.
Torque = Force × Distance
To avoid tipping, the torque generated by the person must not exceed the torque due to the weight of the plank.
Torque due to the person's weight = 445 N × x
Torque due to the plank's weight = 120 N × 1.1 m
445 N × x = 120 N × 1.1 m
4. Now, let's solve for x to find the maximum distance the person can walk on the overhanging part of the plank before it tips.
x = (120 N × 1.1 m) / 445 N
x ≈ 0.298 m
Therefore, the person can walk approximately 0.298 meters on the overhanging part of the plank before it just begins to tip.