Persons A and B are standing on a board of uniform linear density that is balanced on two supports, as shown in the figure. What is the maximum distance x from the right end of the board at which person A can stand without tipping the board? Treat persons A and B as point masses. The mass of person B is twice that of person A, and the mass of the board is half that of person A. Give your answer in terms of L, the length of the board.
Without a diagram, it cannot be figured.
it's not 3/4L or 1/4L
5/8L is the answer for future students who is taking physic class. yw
To find the maximum distance x from the right end of the board at which person A can stand without tipping the board, we need to ensure that the moments or torques on both sides of the supports are balanced. The torque is the product of the force and the perpendicular distance from the pivot point.
Let's analyze the forces acting on the board first:
1. The weight of person A exerting a force downwards at some point on the board.
2. The weight of person B exerting a force downwards at some point on the board.
3. The weight of the board itself.
Since the board is balanced, the total torque must be equal to 0. We can calculate the torque for each force and sum them up:
Torque due to person A = force of person A x distance of person A from the pivot
Torque due to person B = force of person B x distance of person B from the pivot
Torque due to the board = force of the board x distance of the board from the pivot
Now, let's assign some variables:
MA = mass of person A
MB = mass of person B = 2MA (given)
MBD = mass of the board = 0.5MA (given)
LA = distance of person A from the pivot = L - x (since x is from the right end of the board)
LB = distance of person B from the pivot = L (since person B is on the right end of the board)
LD = distance of the board from the pivot = L/2 (since the board is balanced in the center)
We know that the torque is given by the equation:
Torque = Force x Distance
Now let's calculate the torques:
Torque due to person A = (MA)(g)(LA)
Torque due to person B = (2MA)(g)(LB)
Torque due to the board = (0.5MA)(g)(LD)
Adding up the torques:
(MA)(g)(LA) + (2MA)(g)(LB) + (0.5MA)(g)(LD) = 0
Simplifying the equation:
MA(g)(L - x) + 2MA(g)(L) + 0.5MA(g)(L/2) = 0
Now, we can solve for x:
MA(g)(L - x) + 2MA(g)(L) + 0.25MA(g)(L) = 0
(L)(MA)(g) - (MA)(g)(x) + 2(L)(MA)(g) + 0.25(L)(MA)(g) = 0
3.25(L)(MA)(g) - (MA)(g)(x) = 0
(3.25)(L)(MA)(g) = (MA)(g)(x)
3.25(L) = x
So, the maximum distance x from the right end of the board at which person A can stand without tipping the board is 3.25L.
Therefore, the answer is x = 3.25L.