GCSE Balancing Moments made more complex
Balancing Moments made more complex
If you have not read our balancing moments page, please read this one first.
Calculations using Principle of Moments
Force 1, F1 x distance 1, d1 = Force 2, F2 x distance 2, d2
Example calculation
Use the diagram below to answer the following question
Calculate the upward force that is exerted on the drawbridge when it is lifted upwards.
The pivot this time is at the far right and is marked by a circle. Lets draw a diagram that shows both the anticlockwise and clockwise moments.
F1 = 100N, d1 = 0.9m, F2=?N, d2 = 2m
Force 1, F1 x distance 1, d1 = Force 2, F2 x distance 2, d2
100N x 0.9m = ?N x 2m
?N = (100N x 0.9m)/2m
?N = 45N
Example question 2
Use the image below to answer the following question
Calculate the size of the minimum upward force needed to lift the wheelbarrow.
First lets look at the position of both the clockwise and anticlockwise moments.
Ignore the size of the arrows, diagram above is only to ilustrate the idea of both anticlockwise and clockwise moments. The other labels were removed to make the diagram easier to visualise.
Force 1, F1 x distance 1, d1 = Force 2, F2 x distance 2, d2
F1 = 500N, d1 = 0.25m, F2=?N, d2 = 1m
500N x 0.25m = ?N x 1m
?N = (500N x 0.25m)/1m
?N = 125N
1.Use the image below to answer the following questions The weight of the box on the trolley is 500N. The line of action for the weight force is 0.25m from the pivot. A pull force is applied to the handle that is 1.3m from the pivot.
Calculate the size of the pull force F1.
2.Use the image below to help you to answer the following question. Calculate the size of the counter balance force.
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