A student of mass 78 kg wants to walk beyond
the edge of a cliff on a heavy beam of mass
180 kg and length 10 m. The beam is not
attached to the cliff in any way, it simply lays
on the horizontal surface of the clifftop, with
one end sticking out beyond the cliff’s edge:The students want to position the beam so
it sticks out as far as possible beyond the edge,
but he also wants to make sure he can walk to
the beam’s end without falling down.
How far from the edge of the ledge can the
beam extend?
Answer in units of m.
Summing moments is the key.
the cg is five ft from the end, so where is the fulcrum (edge of cliff)?
78*g*d-180g(5-d)=0
solve for d, the distance the beam overhangs.
45
The equation by bobpursley is correct! Make sure to follow order of operations.
78gd+180gd=180g(5)
d=[180g(5)]/(78g+180g)
or you can divide out g
d=(180*5)/(78+180)
To determine how far from the edge of the ledge the beam can extend without the student falling down, we need to consider the torque and balance of forces in the system.
Torque, represented by τ, is the rotational force acting on an object. In this case, the torque exerted by the student's weight on the beam must be balanced by the torque exerted by the weight of the beam itself.
The torque exerted by an object is equal to the product of the force acting on it and the perpendicular distance from the axis of rotation. In this scenario, the force is the weight of the object, and the axis of rotation is the point where the beam extends beyond the edge of the cliff.
Let's denote the distance from the edge of the ledge to the point where the beam extends as x.
Now let's calculate the torque exerted by the student's weight:
τ_student = Force_student * x
The force exerted by the student's weight can be calculated using Newton's second law:
Force_student = mass_student * acceleration_due_to_gravity
Similarly, let's calculate the torque exerted by the beam's weight:
τ_beam = Force_beam * (length_of_beam - x)
The force exerted by the beam's weight can be calculated in the same way:
Force_beam = mass_beam * acceleration_due_to_gravity
For the system to be in equilibrium and prevent the student from falling, the torque exerted by the student's weight must be balanced by the torque exerted by the beam's weight:
τ_student = τ_beam
Substituting the above equations and rearranging:
mass_student * acceleration_due_to_gravity * x = mass_beam * acceleration_due_to_gravity * (length_of_beam - x)
Now, we can solve for x:
mass_student * x = mass_beam * (length_of_beam - x)
Rearranging and simplifying:
mass_student * x = mass_beam * length_of_beam - mass_beam * x
x * (mass_student + mass_beam) = mass_beam * length_of_beam
x = (mass_beam * length_of_beam) / (mass_student + mass_beam)
Plugging in the given values:
mass_student = 78 kg
mass_beam = 180 kg
length_of_beam = 10 m
acceleration_due_to_gravity = 9.8 m/s^2
Calculating the value of x:
x = (180 kg * 10 m) / (78 kg + 180 kg)
x = 1800 kg·m / 258 kg
x ≈ 6.98 m
Therefore, the beam can extend approximately 6.98 meters beyond the edge of the ledge without the student falling down.