A uniform plank of lent light 5m and weight 225N Rests horizontally on two supports with 1.1m of the plank hanging over the right support. To what distance,X, can a person weighing 450 walk On the overhanging part of the plank before it just begins to tip?
Sorry for the typos. It should say, "of length"
The sum of torques has to be zero. So, the sum respectively the reference point (the right support) is
225•1.4 = 450•x
x=225•1.4/450 = 0.7 m
Why does it have to be zero?
The two conditions of equilibrium are:
1. Concurrent Equilibrium : the sum of vector forces through a point is zero.
2. Coplanar equilibrium: the sum of forces in a plane is zero and the sum of the torques around the axis of the plane is zero.
Here we've used the 2nd rule
Should the 1.4 that you used be 1.1?
1.4 is the distance between the center of the plank (its center of mass) and right support.
1.4 = (5/2)-1.1
To determine the maximum distance a person can walk on the overhanging part of the plank before it begins to tip, we need to consider the balance of moments or torques acting on the plank.
First, let's calculate the torque due to the weight of the plank. The weight of the plank acts as a force acting vertically downward at its center of mass, which is at the midpoint of the 5-meter plank. The weight of the plank, W_plank, can be calculated using the formula W = m * g, where m is the mass and g is the acceleration due to gravity.
Given that the weight of the plank is 225 N, we can calculate the mass of the plank as follows:
mass = weight / gravity
mass = 225 N / 9.8 m/s^2
mass ≈ 22.96 kg
Since the plank is uniform, we assume its weight acts at its center of mass, which is located halfway along its length. Therefore, the distance of the center of mass from the left support is 5 m / 2 = 2.5 m.
The torque due to the weight of the plank is given by the formula torque = force * distance. In this case, the force is the weight of the plank and the distance is the distance of the center of mass from the left support. Thus:
torque_plank = weight_plank * distance_center_of_mass
torque_plank = 225 N * 2.5 m
torque_plank = 562.5 N⋅m
To find the maximum distance (X) a person can walk on the overhanging part of the plank without tipping it, we must consider the balance of torques.
When the plank begins to tip, its left end will begin to lift, and there will be a torque acting in the counterclockwise direction at the point where the plank rests on the supports. We can calculate this torque by considering the weight of the person and their distance from the right support.
Let the person's weight be W_person = 450 N. The torque due to the person's weight, torque_person, is given by:
torque_person = weight_person * distance_person
torque_person = 450 N * X
For the plank to just begin to tip, the torques must be balanced. Therefore, equating the torque due to the weight of the plank to the torque due to the person's weight, we have:
torque_plank = torque_person
562.5 N⋅m = 450 N * X
Now we can solve for X:
X = torque_plank / weight_person
X = 562.5 N⋅m / 450 N
X ≈ 1.25 m
Therefore, a person weighing 450 N can walk on the overhanging part of the plank for a maximum distance of approximately 1.25 meters before causing the plank to tip.