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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Explanation:
(a) Displacement of an object is the shortest path covered by it.
In this problem, a student is biking to school. She travels 0.7 km north, then realizes something has fallen out of her bag. She travels 0.3 km south to retrieve her item. She then travels 0.4 mi north to arrive at school.
0.4 miles = 0.64 km
displacement = 0.7-0.3+0.64 = 1.04 km
(b) Average velocity = total displacement/total time
t = 15 min = 0.25 hour
Hence, this is the required solution.
Answer:
Speed of the this part is given as
Also the direction of the velocity of the third part of plate is moving along 135 degree with respect to one part of the moving plate
Explanation:
As we know by the momentum conservation of the system
we will have
here we know that
the momentum of two parts are equal in magnitude but perpendicular to each other
so we will have
now from above equation we have
Also the direction of the velocity of the third part of plate is moving along 135 degree with respect to one part of the moving plate