A 30.0-kg child sits on one end of a long uniform beam having a mass of 20.0 kg, and a 40.0-kg child sits on the other end. The
beam balances when a fulcrum is placed below the beam a distance 1.10 m from the 30.0-kg child. How long is the beam?
1 answer:
let the length of the beam be "L"
from the diagram
AD = length of beam = L
AC = CD = AD/2 = L/2
BC = AC - AB = (L/2) - 1.10
BD = AD - AB = L - 1.10
m = mass of beam = 20 kg
m₁ = mass of child on left end = 30 kg
m₂ = mass of child on right end = 40 kg
using equilibrium of torque about B
(m₁ g) (AB) = (mg) (BC) + (m₂ g) (BD)
30 (1.10) = (20) ((L/2) - 1.10) + (40) (L - 1.10)
L = 1.98 m
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Answer:
The sled required 9.96 s to travel down the slope.
Explanation:
Please, see the figure for a description of the problem. In red are the x and y-components of the gravity force (Fg). Since the y-component of Fg (Fgy) is of equal magnitude as Fn but in the opposite direction, both forces get canceled.
Then, the forces that cause the acceleration of the sled are the force of the wind (Fw), the friction force (Ff) and the x-component of the gravity force (Fgx).
The sum of all these forces make the sled move. Finding the resulting force will allow us to find the acceleration of the sled and, with it, we can find the time the sled travel.
The magnitude of the friction force is calculated as follows:
Ff = μ · Fn
where :
μ = coefficient of kinetic friction
Fn = normal force
The normal force has the same magnitude as the y-component of the gravity force:
Fgy = Fg · cos 30º = m · g · cos 30º
Where
m = mass
g = acceleration due to gravity
Then:
Fgy = m · g · cos 30º = 87.7 kg · 9.8 m/s² · cos 30º
Fgy = 744 N
Then, the magnitude of Fn is also 744 N and the friction force will be:
Ff = μ · Fn = 0.151 · 744 N = 112 N
The x-component of Fg, Fgx, is calculated as follows:
Fgx = Fg · sin 30º = m·g · sin 30º = 87.7 kg · 9.8 m/s² · sin 30º = 430 N
The resulting force, Fr, will be the sum of all these forces:
Fw + Fgx - Ff = Fr
(Notice that forces are vectors and the direction of the friction force is opposite to the other forces, then, it has to be of opposite sign).
Fr = 161 N + 430 N - 112 N = 479 N
With this resulting force, we can calculate the acceleration of the sled:
F = m·a
where:
F = force
m = mass of the object
a = acceleration
Then:
F/m = a
a = 479N/87.7 kg = 5.46 m/s²
The equation for the position of an accelerated object moving in a straight line is as follows:
x = x0 + v0 · t + 1/2 · a · t²
where:
x = position at time t
x0 = initial position
v0 = initial velocity
t = time
a = acceleration
Since the sled starts from rest and the origin of the reference system is located where the sled starts sliding, x0 and v0 = 0.
x = 1/2· a ·t²
Let´s find the time at which the position of the sled is 271 m:
271 m = 1/2 · 5.46 m/s² · t²
2 · 271 m / 5.46 m/s² = t²
<u>t = 9.96 s </u>
The sled required almost 10 s to travel down the slope.
