Answer:
A 20 N force acting downward on the right end of the lever would keep it horizontal.
Explanation:
Recall that in order to keep the lever horizontal and without rotating, one needs the sum of the acting torques to be zero.
The torque that the 60 N force of the hanging block produces, contributes to a positive torque (according to the convention counter-clock motion is positive), and its magnitude is the product of the force times its distance to the fulcrum: 60 N * 1 m = 60 Nm.
So we need a negative torque of the same absolute magnitude from the right hand side. Then a force downward will accomplish the negative sign required (according to convention clockwise motion is negative). The magnitude of that torque should equal 60 Nm as well, so we find which force, acting at the other end (3 m from the fulcrum) would accomplish such:
[tex]60\, N\,m\,= F\,*\,3 \,m\\F=60/3 \,\,N\\F=20\,N[/tex]
Therefore, a 20 N force acting downward would keep the lever horizontal.
