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A student of mas 70kg wants to walk beyond the edge of a cliff on a heavy beam of mass 300kg and length 10m. The beam is not at...Asked by Ashley
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.
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.
Answers
Answered by
bobpursley
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.
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.
Answered by
Anonymous
45
Answered by
nope
The method posted by bobpursley is wrong
Answered by
Yes!
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)
78gd+180gd=180g(5)
d=[180g(5)]/(78g+180g)
or you can divide out g
d=(180*5)/(78+180)
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