2010 P1 Question 5.

A load is to moved using a wheelbarrow.The total mass of the load and wheelbarrow is 60kg.The gravitational field strength is 10N/kg.What is the size of the force needed to just lift the loaded wheelbarrow?

To solve such questions,we shall base our knowledge on the priciple of moments which says that clockwise moments are equal to anti-clockwise moments when the system is at equilibrium or balanced.The other thing to take note of is the class from which this simple machine comes from,we can say this one is from the third class which has its load in between the pivot and the effort.This kind of simple machine is of the same class as the fishing rod.

STEP 1.

Sketch the diagram in simple 2D non-graphical form if it was presented such as this one.This helps manipulate the diagram to get the data you want.

STEP 2.

Look for available turning effects of the given forces by knowing which force can act perpendicularly to the horizontal or vertical distance..In our case,we have two that can act perpendicularly to the distance and that if from the pivot to the force and from the pivot to the center of mass of the wheelbarrow.

STEP 3.

Collect your data.

F₁=60 × 10=600N

Distance₁=70cm=0.7m

F₂=F

Distance₂=70+50=120=1.2m

STEP 4.

Calculate by first stating the formula.

moment₁=Distance₁ × force ₁

moment₁=600 × 0.7

moment₁=420Nm.

moment₂=Distance₂ × force ₂

moment₂=1.2 × F

moment₂=1.2F Nm.

moment₁ = moment₂

420=1.2F

420/1.2=1.2F/1.2

350=F

F=350N