The diagram shows a student using a screwdriver to lift the lid from a paint tin. The rim of the tin is the pivot. The student pushes down on the handle wit...
Question 1 Report
The diagram shows a student using a screwdriver to lift the lid from a paint tin. The rim of the tin is the pivot. The student pushes down on the handle with 60 N. The distance from the pivot to the hand is 0.18 m. The screwdriver blade touches the lid 0.015 m from the pivot. The lid is stuck because dried paint forms a seal around its edge.
The forces are perpendicular to the screwdriver.
(a) What is the name of this simple machine? [1]
(b) Calculate the moment from the student's hand. [2]
(c) Calculate the upward force on the lid. [2]
Fig. 1 shows a laboratory test of a metre rule used as a lever. A 2.0 N load hangs from the 20 cm mark. The rule is supported at the 50 cm mark. A spring balance pulls down at the 80 cm mark until the rule is horizontal. The student records the force shown by the spring balance. The weight of the rule is ignored in this simple model.
The distances from the pivot are measured along the rule.
(a) What is the distance of the load from the pivot? [1]
(b) Calculate the moment of the load about the pivot. [1]
(c) Calculate the spring-balance force needed for balance. [2]
Answer Details
First lever system
- (a) The screwdriver is a lever: it turns about a pivot to multiply the force. [1]
- (b) Moment = force × perpendicular distance from pivot.
\[M=60\text{ N}\times0.18\text{ m}=10.8\text{ N m}\]Moment from the hand = 10.8 N m. [2] - (c) For the screwdriver just to lift the lid, the turning moments balance:
\[F\times0.015\text{ m}=10.8\text{ N m}\]
\[F=\frac{10.8}{0.015}=720\text{ N}\]The upward force on the lid is 720 N. The much shorter distance on the lid side produces a much larger force. [2]
Metre-rule lever system
- (a) The load is at 20 cm and the pivot is at 50 cm, so its distance from the pivot is \(50-20=30\) cm, or 0.30 m. [1]
- (b) \[M=2.0\text{ N}\times0.30\text{ m}=0.60\text{ N m}\]Moment of the load = 0.60 N m. [1]
- (c) The spring balance is at 80 cm, which is \(80-50=30\) cm = 0.30 m from the pivot. Its moment must equal 0.60 N m:
\[F=\frac{0.60\text{ N m}}{0.30\text{ m}}=2.0\text{ N}\]Spring-balance force = 2.0 N. [2]
Exam reminder: use perpendicular distance in metres, and equate clockwise and anticlockwise moments when a lever is balanced.
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