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  1. Problems
  2. /A crowbar 1.5 m long rests on a stone 0.25 m from the end under a 900 N load…
EasyPhysicsHigh School (11–12)Mechanics › Rotational MotionNumericalBoard ExamAPA-LevelIB

A crowbar 1.5 m long rests on a stone 0.25 m from the end under a 900 N load…

StemPy LabContributed by StemPy Lab

Try it on paper first. A hint, the key idea and the worked solution are here when you need them.

A crowbar 1.5 m1.5\ \text{m}1.5 m long rests on a stone 0.25 m0.25\ \text{m}0.25 m from the end under a 900 N900\ \text{N}900 N load.

  1. What effort, at right angles to the bar at its other end, balances the load?
  2. What is the mechanical advantage?

Check your answer

Work it out on paper, then type your number. The answer itself stays hidden.

N
Need a reminder? The key ideaShowHide

A rigid body is in equilibrium when the net force is zero and the net torque about every point is zero. Take torques about a point where an unknown force acts, so that it drops out.

The worked solution

Have you tried it?

Work it out on paper first. It is fine to get stuck or make a mistake: that is how we learn. When you are ready, go through the solution one step at a time, and check each step against your own.

Plan. The stone is the pivot.

Step 1. (a) 900 N×0.25 m=E×1.25 m900\ \text{N} \times 0.25\ \text{m} = E \times 1.25\ \text{m}900 N×0.25 m=E×1.25 m, so E=180 NE = 180\ \text{N}E=180 N.

Step 2. (b) loadeffort=900 N180 N=5\dfrac{\text{load}}{\text{effort}} = \dfrac{900\ \text{N}}{180\ \text{N}} = 5effortload​=180 N900 N​=5.

Answer: (a) 180 N180\ \text{N}180 N. (b) 555.

Check. The arms are in the ratio 1.25:0.25=5:11.25 : 0.25 = 5 : 11.25:0.25=5:1, the same as the forces the other way round.

Watch out. The effort's moment arm is 1.5 m−0.25 m=1.25 m1.5\ \text{m} - 0.25\ \text{m} = 1.25\ \text{m}1.5 m−0.25 m=1.25 m, not the bar's whole length.

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