Question

Could someone please help me figure out this question

Answer 1

1. mass should be placed 0.4 m to the left of the pivot
2. T = 68.6 N

Step-by-step explanation:

After adding the 5kg mass, in order for the rod to remain horizontal → ∑τ = 0

τ of the rod is clockwise, therefore the mass has to be placed on the left side from the pivot to create that is anti clockwise

`\Sigma\tau=0`

`\tau_(rod)+\tau{mass}=0`

`-r_(rod)\cdot w_(rod)+r_(mass)\cdot w_(mass)=0`

`-1*2*9.8+r_(mass)*5* 9.8=0`

`5r_(mass)=1*2`

`r_(mass)=0.4\ m` (mass is 0.4 m to the left of the pivot)

`\Sigma F=0`

`T-w_(rod)-w_(mass)= 0`

`T=2*9.8+5*9.8`

`=68.6\ N`