Moments
Q1.
(a) Define the moment of a force about a point.
![](syllabus.gif)
A moment is the force multiplied by the perpendicular distance
from the (pivot) point to the line of action of the force.![](../../../../graphics/symbols/nuclides/ticksmall.png)
[2 marks]
(b) The diagram shows a gripper which is used for hand strengthening exercises.
![](gripper.gif)
The diagram shows the gripper being squeezed.
In this situation, the gripper is in equilibrium.
The force produced by the fingers is equivalent to the single force X of magnitude 250 N acting in the direction shown in the diagram.
A force, Y, is exerted by the spring which obeys Hooke's law.
(i) Calculate the moment of force X about the pivot. State an appropriate unit.
![](../../../../graphics/equations/dataSheet/moment.png)
d = 48 x 10-3m
F = 250 N
moment = Fd = 250 x 48 x 10-3 = 12
Nm ![](../../../../graphics/symbols/nuclides/ticksmall.png)
[2 marks]
(ii) Calculate force Y.
d = 27 x 10-3m
F = ? N
12 = F x 27 x 10-3
F = 12/(27 x 10-3)
= 444 N
F = 440 N ![](../../../../graphics/symbols/nuclides/ticksmall.png)
[2 marks]
(iii) The extension of the spring is 15 mm. Calculate the spring constant k of the spring. Give your answer in N m–1.
![](../../../../graphics/equations/dataSheet/materials.png)
F = kΔl
444 = k x 15 x 10-3
k = 444/(15 x 10-3) ![](../../../../graphics/symbols/nuclides/ticksmall.png)
k = 29600 N/m
k = 3.0 x 104 Nm-1 ![](../../../../graphics/symbols/nuclides/ticksmall.png)
[2 marks]
(iv) Calculate the work done on the spring to squeeze it to the position shown in the diagram.
E = 1/2 FΔl
E = 0.5 x 444 x 15 x 10-3 ![](../../../../graphics/symbols/nuclides/ticksmall.png)
E = 3.3 J ![](../../../../graphics/symbols/nuclides/ticksmall.png)
[2 marks]
(Total 10 marks)