A uniform plank of length 5.5 m and weight 219 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 447 N walk on the overhanging part of the plank before it just begins to tip?
Assume the weight of the plank acts through the denter of mass. When tipping starts. there will be no force applied at the left support, and the net torque about the right support (the pivot point) will be zero. Write that as an equation and solve for the distance from the picot point.
Well, isn't this a balancing act! If we want to find the distance from the pivot point at which the plank begins to tip, we need to consider the torques acting on it.
First, let's determine the weight of the plank. Since weight = mass x gravity, and we know the weight is 219 N, we can solve for mass. Assuming gravity is 9.8 m/s^2, we have mass = 219 N / 9.8 m/s^2.
Now, let's consider the torques acting on the plank. We have the weight of the plank acting downward at its center of mass, and the weight of the person acting downward at a certain distance x from the pivot point.
Since the plank is in equilibrium, the sum of these torques must be zero. The torque due to the weight of the plank is (1/2)(5.5 m)(g)(5.5 m / 2), where g is the acceleration due to gravity. The torque due to the weight of the person is (x)(447 N). Setting these two torques equal to each other, we have:
(1/2)(5.5 m)(g)(5.5 m / 2) = (x)(447 N)
Now, if you're anything like me, equations can be a bit dry. So, let's add a touch of humor to make things more fun!
If the equation was a joke, it would go something like this:
Why did the plank hire a mathematician? Because it needed help with all the torque-y business! *ba dum tss*
But fear not, my friend, solving this equation is no laughing matter! By plugging in the appropriate values and solving for x, you'll find the distance at which the plank just begins to tip.
And remember, when in doubt, always seek balance in life - both literally and metaphorically!
To determine the distance from the pivot point where the plank will just begin to tip, we need to consider the torques acting on the plank.
The weight of the plank can be assumed to act through its center of mass, which is at the midpoint of the plank. The weight of the person will also create a torque around the pivot point as they move along the plank.
Let's denote the distance from the pivot point to the point where the person is standing as "x" (measured from the right end of the plank), and the distance from the pivot point to the center of mass of the plank as "d" (half the length of the plank).
The torque due to the weight of the person is given by the equation:
Torque_person = person_weight * x
The torque due to the weight of the plank can be calculated as:
Torque_plank = plank_weight * d
Since there is no net torque acting on the system when tipping starts, we can set up the equation:
Torque_person = Torque_plank
person_weight * x = plank_weight * d
Substituting the given values:
447 N * x = 219 N * (5.5 m / 2)
447 N * x = 1199.25 N*m
Now, we can solve for x:
x = 1199.25 N*m / 447 N
x ≈ 2.682 m
Therefore, the person can walk a distance of approximately 2.682 meters on the overhanging part of the plank before it just begins to tip.
To solve this problem, we need to analyze the torque acting on the plank and set it equal to zero when tipping starts.
Let's start by calculating the weight of the plank. The weight of the plank is given as 219 N, which means the weight acts downward towards the center of the earth.
Since the weight of an object can be modeled as acting through its center of mass, we can assume that the weight of the plank acts through its center of mass, which is at the midpoint of the plank.
Now let's consider the forces and torques acting on the plank when a person walks on the overhanging part. We have the weight of the plank acting downward at the center of mass, the weight of the person acting downward at the distance x from the right support, and the reaction forces acting upward at the two supports.
The net torque acting on the plank about the right support (the pivot point) will be zero when tipping starts. This means that the torques from the weight of the plank and the weight of the person must balance.
The torque due to the weight of the plank is given by T(plank) = Weight(plank) * distance from pivot point
The torque due to the weight of the person is given by T(person) = Weight(person) * distance from pivot point
Setting these torques equal to each other, we have:
Weight(plank) * distance from pivot point = Weight(person) * distance from pivot point
219 N * 1.1 m = 447 N * x
Using this equation, we can solve for the distance x:
x = (219 N * 1.1 m) / 447 N
Calculating this expression, we find:
x ≈ 0.537 m
Therefore, a person who weighs 447 N can walk approximately 0.537 m on the overhanging part of the plank before it just begins to tip.