A uniform plank of length 5.0 m and weight 225 N rests horizontally on two supports, with 1.1 m of the plank hanging over the right support (see the drawing). To what distance x can a person who weighs 404 N walk on the overhanging part of the plank before it just begins to tip?
(225)(5.0) = (1.1)(400x) = 2.55m
Is this correct? Thank you!
When the person reaches the maximum distance x along the overhanging
part, the plank is just about to rotate about the right support.
At that instant, the plank loses contact with the left support, which consequently exerts no force on it. This leaves only three vertical forces acting on the plank: the weight W of the plank, the force FR due to the right support, and the force P due to the person. The force FR acts at the right support, which we take as the axis, so its lever arm is zero.
225•(3.9-2.5) = 404•x
x=225•1.4/404=0.78 m
A 15 ft plank of mass 50 kg is laid across two ladders. The ladders are 5 ft apart, and the plank is laid such that 5 ft of it hang off at each end. You have a mass of 90 kg.
(a) How far out past the ladder can you walk before it gets dangerous?
You have a friend of mass 50 kg who is standing on the plank between the ladders, 2.2 ft from the left ladder.
(b) How far out past the rightmost ladder can you safely walk now?
No, the calculation you provided is not correct. To determine the maximum distance a person can walk before the plank begins to tip, you need to consider the principle of torques.
The torque exerted by an object is equal to the product of its weight (W) and the distance from the pivot point (r) at which it acts, and it must be balanced for the system to be in equilibrium.
In this case, the weight of the plank is acting at its center, which is 2.5 m from each end. The weight of the person will act at a distance of (5.0 - 1.1 - x) from the left end of the plank. Assuming the person is at the very end of the overhanging part, the distance would be (5.0 - 1.1 - x). The weight of the person will cause a torque in the clockwise direction.
To balance the torques, the torque exerted by the person must be equal to the torque exerted by the plank. Therefore, we can set up the equation:
(225 N)(2.5 m) = (404 N)(5.0 - 1.1 - x)
Simplifying:
562.5 N*m = 2916 N - 404 N*x - 444.4 N
Now solve for x:
404 N*x = 2916 N - 444.4 N - 562.5 N*m
404 N*x = 2009.1 N - 562.5 N*m
x = (2009.1 N - 562.5 N*m) / 404 N
x = 4.38 m
Therefore, the person can walk up to a distance of approximately 4.38 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 begins to tip, we need to consider the torque and equilibrium conditions.
In this case, torque is the force applied by the person walking on the plank, causing it to tip. The torque is calculated by multiplying the force applied by the person (404 N) by the distance from the left support to the point where the person is walking (x).
The torque exerted by the weight of the plank is given by multiplying the weight of the plank (225 N) by the distance from the left support to the center of the plank (5.0/2 = 2.5 m).
To maintain equilibrium, the sum of the torques must be zero. Hence, we can set up the equation:
(Torque exerted by the person) + (Torque exerted by the weight of the plank) = 0
(404 N)(x) + (225 N)(2.5 m) = 0
Simplifying the equation, we have:
404x + 562.5 = 0
Solving for x:
404x = -562.5
x = -562.5 / 404
However, since x represents a distance, it cannot be negative. Hence, the maximum distance the person can walk on the overhanging part of the plank before it begins to tip is:
x = 562.5 / 404
Calculating the value:
x ≈ 1.39 m
Therefore, the correct maximum distance a person weighing 404 N can walk on the overhanging part of the plank before it just begins to tip is approximately 1.39 m.