A uniform plank of length 6.0 m and weight 231 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 451 N walk on the overhanging part of the plank before it just begins to tip?
You have two equations you can write.
1) Sum forces vertical=0
LS + RS-231-451=0 where LS, Rs are the forces on the right support.
2) You can sum moments about any point. Hmmm. YOu didn't say where the left support is.
Anyway, sum momemts about the left support (RS, cg, man) and set to zero.
Then sum moments about the RS.
That ought to give you sufficent equations to solve.
To find the maximum distance x that a person can walk on the overhanging part of the plank before it begins to tip, we need to consider the balance of torques.
First, let's consider the torques acting on the plank. The weight of the plank itself (231 N) can be considered as acting at its midpoint, which is 3.0 m from each support.
Next, the weight of the person (451 N) can be considered as acting at the position where they are standing on the plank. Let's call this position x.
To prevent the plank from tipping, the total torque due to the person's weight must be balanced by the torque due to the weight of the plank. Mathematically, this can be expressed as:
451 N * x = 231 N * (6.0 m - x)
Simplifying the equation, we get:
451x = 1386 - 231x
Combining like terms:
682x = 1386
Dividing both sides by 682:
x = 2.03 m
Therefore, a person who weighs 451 N can walk on the overhanging part of the plank for a maximum distance of approximately 2.03 meters before it just begins to tip.