What Position Does the Woman Reach When the Beam Tips?

[mandy9008]

Homework Statement

A beam resting on two pivots has a length of L = 6.00 m and mass M = 72.0 kg. The pivot under the left end exerts a normal force n1 on the beam, and the second pivot placed a distance l = 4.00 m from the left end exerts a normal force n2. A woman of mass m = 51.1 kg steps onto the left end of the beam and begins walking to the right as in the figure below. The goal is to find the woman's position when the beam begins to tip.

[PLAIN]http://www.webassign.net/sercp8/p8-12.gif[/IMG

a. What is n1 when the beam is about to tip?
b. Use the force equation of equilibrium to find the value of n2 when the beam is about to tip.
c. Using the result of part (a) and the torque equilibrium equation, with torques computed around the second pivot point, find the woman's position when the beam is about to tip.

The Attempt at a Solution

a. n1=0 N
b. n2=mg
n2= (51.1kg)(9.8 m/s²)
n2=500.78 N
c. since F is 0 in part a, I would think that this answer will be zero as well

 

[zhermes]

You have to include the mass of the beam itself as-well.

 

[mandy9008]

okay, so n2 will be 1206.38 instead, assuming that you meant: n2= (51.1kg + 72.0 kg)(9.8 m/s2). so what do i need to do for part c?

 

[zhermes]

That looks about right. Whats the equation for torque? And the instant before it begins to tip, what is the net torque? It might help if you draw a picture of this situation, consider what torques are in what direction, remember to include the beam.

 

[rwisz]

.

I would like to clarify a few things about the problem and the proposed solution. Firstly, the concept of equilibrium refers to a state in which all forces and torques acting on an object are balanced, resulting in no net acceleration or rotation. In this problem, we are dealing with static equilibrium, meaning the object is not moving or rotating.

Now, let's address the solution provided. In part a, it is stated that n1=0 N. However, this is not entirely correct. Since the woman is standing on the beam, there will be a normal force exerted by the beam on her, which will in turn exert a normal force on the pivot. This normal force will counteract the weight of the beam and woman, and thus n1 will not be equal to 0 N. To find the value of n1, we can use the force equation of equilibrium, which states that the sum of all forces in the x-direction must equal 0. This can be written as:

n1 + n2 - mg = 0

Solving for n1, we get:

n1 = mg - n2

Substituting in the given values, we get:

n1 = (72.0 kg)(9.8 m/s^2) - (51.1 kg)(9.8 m/s^2) = 216.6 N

Therefore, the correct answer for part a is n1 = 216.6 N.

Moving on to part b, the solution provided is correct. We can use the same force equation of equilibrium to solve for n2:

n1 + n2 - mg = 0

Substituting in the given values and solving for n2, we get:

n2 = mg - n1 = (51.1 kg)(9.8 m/s^2) - (216.6 N) = 284.08 N

So, the correct answer for part b is n2 = 284.08 N.

Finally, in part c, we are asked to find the woman's position when the beam is about to tip. To do this, we can use the torque equilibrium equation, which states that the sum of all torques acting on an object must equal 0. This can be written as:

Στ = 0

In this case, we will take torques around the second pivot point, since we are given the distance l from