A worker tearing down a barn uses a crowbar 4feet long with the shorter end
being 2 inches from the fulcrum.
If the worker can exert a force of 150lbs at one end of the crowbar, how great a
force can be exerted at the
other end?
(Note the differences in units, 2 inches and 4feet. Units must be the same when
using the lever formula)
He can exert _________ lbs of force.
Formula: (w1)(d1)=(w2)(d2)
4ft = 48in
Work is the same on both sides, and W = Fd
W1 = W2
F1*d1 = F2*d2
150lbs*48in = F2*2in
F2 = 3600lbs
I believe the pubilc is more right than the statistics reported here. Unfortunately, these numbers don't have much bearing on the quality of education. I'm very sour on the state of our pubilc education. The system is in very bad need of serious competition.
To find the force that can be exerted at the other end of the crowbar, we can use the lever formula: (w1)(d1) = (w2)(d2), where w1 and w2 are the weights or forces and d1 and d2 are the corresponding distances from the fulcrum.
Given:
w1 = 150 lbs (the force exerted by the worker)
d1 = 2 inches (the distance from the fulcrum to the shorter end)
We need to convert the units to be consistent before using the formula. Since the formula requires the units to be the same, let's convert the 2 inches to feet.
There are 12 inches in 1 foot, so 2 inches is equal to (2/12) = 1/6 foot.
Now we have:
w1 = 150 lbs
d1 = 1/6 foot
Next, we need to find the values of w2 and d2, the force and distance at the other end of the crowbar.
Since the crowbar is a rigid object, the sum of moments (or torques) on both sides of the fulcrum should be equal. That means (w1)(d1) = (w2)(d2).
Plugging in the given values:
(150 lbs)(1/6 foot) = (w2)(4 feet)
Now we can solve for w2:
(150/6) = w2(4)
w2 = (150/6)/(4)
w2 = 25 lbs
Therefore, the worker can exert a force of 25 lbs at the other end of the crowbar.