Question

A uniform 0.093 kg meter stick is supported at the 15 cm and 92 cm marks. When a 0.250 kg object is placed at the 6.0 cm mark, what are the magnitudes of the forces supporting the meter stick?

Answer 1

F1 = 3.23 N

F2 = 0.128 N

Step-by-step explanation:

Set the point of rotation at the 15 cm mark, where F1 is.

The meter stick is not moving, so it is in equilibrium.

Let s= meter stick, o = the object, F1 and F2 are the supporting forces

Sum of the forces in the y-direction:

∑Fy = F1 + F2 -Ws - Wo = ma = m(0) = 0

-(.25kg)(9.8 m/s2) - (0.25 kg)(9.8 m/s2) + F1 + F2 = 0

τ = torque = rFsinθ

sin90 = 1

∑τ = τo + τ1 - τs + τ2 = I(∝) = I(0) = 0

CCW rotation is +

Now solve for F2:

(0.09m)(0.25 kg)(9.8 m/s2) - (0.35m)(0.093 kg)(9.8 m/s2) + (0.77m)(F2) = 0

you have to figure out the r values based on using the 15 cm mark as the point of rotation

0.22 - 0.319 = (-.77)F2 I left the units out--that's a lot of typing!

F2 = 0.128 N

Now plug the value of F2 into the bolded equation at the top and solve for F1:

-(.25kg)(9.8 m/s2) - (0.25 kg)(9.8 m/s2) + F1 + 0.128 N = 0

F1 = 3.23 N