Question

There is a lever. The fulcrum is 1 m from the left end, and 3 m from the right end. A block of 60 N is hung to the left end. To keep the lever in horizontal, how much force is needed on the right end?

Answer 1

Answer:

20N

Explanation:

From moment of a force at equilibrium.

The clockwise moment is same as the anticlockwise moment.

An anticlockwise moment is ensued by the 60N force and the unknown force F ensues a clockwise moment

60×1 = 3×F

F=60/3= 20N

{ Moment is defined as the force multiplied by the distance of the force from the support}. We see that the unknown force has to be applied at the right end and should be directed downwards.

Answer 2

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:

$60\, N\,m\,= F\,*\,3 \,m\\F=60/3 \,\,N\\F=20\,N$

Therefore, a 20 N force acting downward would keep the lever horizontal.