A 20.0 m long uniform beam weighing 800 N rests on walls A and B, as shown in the figure below.(a) Find the maximum weight of a person who can walk to the extreme end D without tipping the beam.(b) Find the forces that the walls A and B exert on the beam when the same person is standing at point D.

(c) Find the forces that the walls A and B exert on the beam when the same person is standing at a point 2.0 m to the right of B.
(d) Find the forces that the walls A and B exert on the beam when the same person is standing 2.0 m to the right of A.

I have the first two parts done but I don't really know what equations I should be using for the last two parts. my answer for part A was 800N and my answers for part B were 1600N for wall B and 0N for wall A. like I said I just don't know what equations to use, I don't necessarily need someone to go through the entire process of solving the equation.

Well you've got no picture but lets see what we can do description-wise.

a) Sum of torques must be zero (I'm assuming B is between center of board and man). Mg of the board times distance to middle of board to B = weight of man times distance to B
b) Sum of forces must be zero. Your answer looks ok as A should be zero.
c) Now sum torques around A
Clockwise:
weight of board times distance from a to center of board + weight of man times distance from A to man
Counter clockwise:
Normal force from B times distance from A to B. To find normal at A, sum of forces = zero
d)Same thing. sum torques, find A; sum forces, find B

Well, I'm no physics expert, but let's give it a shot with a touch of humor!

To tackle parts (c) and (d), we have to consider the concept of torque. Think of it as a twisty force that can cause rotation. Now, remember that torque depends on the distance from the pivot point. And since we're all about humor, we'll call that distance the "funny twisty radius."

In part (c), when the person is standing 2.0 m to the right of B, the funny twisty radius for wall A is shorter than in part (b). Why? Because the person is closer to B. So, the force exerted by A will be less since it has less leverage to twist the beam. It's like trying to open a jar of pickles with a smaller-sized pickle jar opener. Useless!

Now, in part (d), with the person standing 2.0 m to the right of A, the funny twisty radius for wall B is shorter. This means that wall B has less leverage, and it'll be like trying to open a stuck jar of pickles with a slippery hand covered in pickle juice. Slippery and weak!

So, in short:

(c) Expect wall A's force to be less than in part (b), while wall B's force remains the same. It's like trying to twist open a jar of pickles with a tiny pickle jar opener and a strong grip.

(d) Wall B's force will be less than in part (b), while wall A's force remains unchanged. It's like trying to twist open a jar of pickles with a strong pickle jar opener but a hand weakened from too much laughter.

Remember, take my hilarious explanations with a grain of salt, and consult a physics expert for more accurate answers.

To solve parts (c) and (d), you need to consider the torque exerted by the person's weight on the beam. Torque is the product of a force and the distance from the pivot point (in this case, the end of the beam) to the line of action of that force.

For part (c), where the person is standing at a point 2.0 m to the right of B, the torque exerted by the person's weight must be balanced by the torques exerted by the forces exerted by walls A and B. Since the person's weight is acting downwards, the reaction force from wall A must exert an upward force to create a torque in the opposite direction.

To calculate the forces exerted by the walls, you can use the equation:

Sum of torques = Torque from weight

Wall A exerts a torque with a magnitude of (force from wall A) * (distance of wall A from the pivot point), and wall B exerts a torque with a magnitude of (force from wall B) * (distance of wall B from the pivot point).

For part (c), the torque from the person's weight is (Weight of person) * (distance of person from the pivot point). You can set up the equation and solve for the forces from wall A and wall B.

For part (d), where the person is standing 2.0 m to the right of A, the process is similar, but you will use the new distance for wall A.

Remember that a positive torque causes a clockwise rotation, and a negative torque causes a counterclockwise rotation.

To solve parts (c) and (d), you need to understand the concept of torques and equilibrium. Torque is a rotational force, and in this case, it causes the beam to rotate around a pivot point (probably at wall B). Equilibrium occurs when all the forces and torques acting on an object balance out, resulting in no net force or torque.

Let's start with part (c), where the person is standing at a point 2.0 m to the right of wall B. To find the forces exerted by walls A and B, we need to consider the torques involved.

1. Calculate the torque exerted by the weight of the beam.
- The weight acts downward at the center of the beam (at a distance of 10.0 m).
- The torque can be calculated as T_beam = weight of the beam x distance.
- T_beam = 800 N x 10.0 m.
- T_beam = 8000 Nm.

2. Calculate the torque exerted by the person's weight.
- The person's weight acts downward at point D (at a distance of 12.0 m).
- Let's assume the person weight is W.
- The torque can be calculated as T_person = W x distance.
- T_person = W x 12.0 m.

3. Set up an equation based on equilibrium.
- In equilibrium, the sum of the torques must be zero.
- T_beam + T_person = 0.
- 8000 Nm + W x 12.0 m = 0.

4. Solve the equation to find the person's weight.
- W x 12.0 m = -8000 Nm.
- W = -8000 Nm / 12.0 m.
- W ≈ -667 N.

Since weight is a positive scalar quantity, the negative sign indicates that the person's weight acts in the opposite direction (upwards).

For part (d), when the person is standing 2.0 m to the right of A, the process is similar, but you need to consider the torques differently based on the person's position relative to wall A.

Hope this helps you solve the last two parts of the problem!