Moment of force, rod with mass

[Memo]

Homework Statement

See the photo below. A uniform rod AC with a mass of 0.4kg, hang m1=5kg. Find m2 for the system to be in equilibrium.
Answer hint: m2=2.4(kg)

Relevant Equations

M=F*d*sina

OA*P_A = OH*P + OB*P_B
⇔ OA*P_A = ¾*OA*P + 2OA*P_B
⇔ 5*10 = ¾*0.4*10 + 2*P_B
→P_B=23.5 →m2=2.35 (kg)

My result doesn't match the answer hint, and when I change the rod's centre of mass to OH=1, it fits the answer hint. How do I determine the centre of mass in this situation?

 

[kuruman]

Your answer matches the hint to 2 significant figures which the accuracy of the calculation. Your method is correct.

 

[Memo]

Your answer matches the hint to 2 significant figures which the accuracy of the calculation. Your method is correct.

Thanks for reaching out. Normally, that would be considered incorrect. I meant, the figure gives out nicely, why would my teacher round it.

 

[kuruman]

Thanks for reaching out. Normally, that would be considered incorrect. I meant, the figure gives out nicely, why would my teacher round it.

Because you have to eyeball the distances and the accuracy of the calculations does warrant anything more than two sig figs. Ask your teacher and see what answer you get.

 

[haruspex]

The details of the diagram make me suspicious. The labels O, H, P and the down arrow at P seem to have been added later. The vertical divisions on the rod are unevenly spaced.
I suspect it has been inaccurately recycled.

 

[Lnewqban]

How do I determine the centre of mass in this situation?

The center of mass is located 3.5 length units from each end of the rod.
The mass is evenly distributed along the rod because:
"A uniform rod AC with a mass of 0.4kg".

The answer hint is simply rounding 2.35 kg to 2.40 kg.
The error is only 2.1 %.