Question

Fan and medal c:
a 20 kg stack of bricks is placed one meter on the left of the fulcrum on a 1st class lever. Where would a 5 kg stack of bricks need to placed in order to balance the lever?
A
1 meter to the right of the fulcrum
B
2 meters to the right of the fulcrum.
C
3 meters to the right of the fulcrum.
D
4 meters to the right of the fulcrum.

Answer 1

I think I can solve this pretty easily. First of all, this equation is referring to torque/moment about a point!
We see that when we place an object on one end, it's gravity is going to push that object down with a certain force with respect to its mass. Therefore, we have this situation:
We already know that we want this lever to be completely balanced. In other words, the torque produced on the left side must be the same as the torque produced on the right side! Great, now we know where we need to go!

\[\huge \tau = \vec{F} \times \vec{d}=|\vec{F}||\vec{d}| \sin(\theta)\]
Where torque is the cross product of the force and position vectors. Here we can see that the load is immediately applied vertically to the horizontal lever, so the angle between the force and distance is 90. \(\sin(90)=1\). Additionally, we're not dealing with vectors, only magnitudes, so we can drop the magnitude symbols. Therefore, our final equation is

\[\huge \tau = Fd\]
So now we only have one issue -- we're given mass! This is a simple fix. To find the forces due to gravity (aka, the weight) we must multiply by the gravitational constant, 9.8m/s^2!

\[F_1=9.8m/s^2 \times 20kg=196\text{N}\]\[F_2=9.8m/s^2 \times 5kg=49\text{N}\]

Now, we already said that the torques on both sides must be equal. Therefore,
\[\huge \tau_L=\tau_R\]\[\huge Fd=Fd \implies 196N(1m)=49N(d)\]

So d=??