Physics, asked by charmingdoĺl225, 1 year ago

A uniform metre rule of weight 10 gf is pivoted at 0 Mark. 1. what moment of force depresses the rule?2.how can it be made horizontal by applying least force


NeneAmaano: What's the force?
charmingdoĺl225: torque or couple
12gj45f: 10gf
charmingdoĺl225: how ???
charmingdoĺl225: can you explain
charmingdoĺl225: pls answer fast
NeneAmaano: i answered but sorry that I was late

Answers

Answered by arjunyt13
231

check out the image for answer...it is 100% correct

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Answered by mindfulmaisel
63

Magnitude of force = 5 gf .

Given:

The given data suggests that the uniform meter rule of weight 10gf is pivoted at zero mark.  

Solution:

So the moment of force is given by the product of magnitude of force and perpendicular distance corresponding to the force action line in relation to rotational axis.  

Thereby the moment of force = 50 \mathrm{cm} \times 10 \mathrm{gf} = 500 gf\ cm.  

And therefore the least force applied will have the perpendicular distance at the highest, therefore we have the distance of 100 cm mark highest where the force application be least.  

\begin{array}{l}{\text {Moment of force = Magnitude of force} \times \text { Mark distance }} \\ {500=\text { Force magnitude } \times 100}\end{array}

Magnitude of force = 5 gf which is to be applied upwards.

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