A man drags a 71-kg crate across the floor at a constant velocity by pulling on a strap attached to the bottom of the crate. The
1 answer:
Answer:
T = 540 N (to two significant digits)
Explanation:
Let the crate dimension L be from strap attachment to floor contact
Let T be the strap tension
sum moments about the floor contact point to zero
mg[½Lcos25] - Tsin61[Lcos25] + Tcos61[Lsin25] = 0
L is common to all terms, so divides out.
½(71)(9.8)cos25 = T(sin61cos25 - cos61sin25)
T = (71)(9.8)cos25 / (2(sin61cos25 - cos61sin25))
T = 536.428020...
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Answer:
(-1.5,-1.5)m
Explanation:
we know that:
where is the location of the center of gravity in the axis x,
is the mass of the object i and
the first coordinate of center of gravity of object i.
so:
Where is the first coordinate of the center of gravity for the fourth object.
Therefore, solving for , we get:
At the same way:
where is the location of the center of gravity in the axis y,
is the mass of the object i and
the second coordinate of center of gravity of object i. replacing values we get:
Where is the second coordinate of the center of gravity for the fourth object.
solving for :
It means that the object of mass 8kg have to be placed in the
coordinates (-1.5,-1.5) m.