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A uniform rod of length 1 m mass 4 kg is supports on tow knife-edges placed10 cm from each end. A 60 N weight is suspended at 30 cm from one end. The reactions at the knife edges is.

Answer» <html><body><p>a. 60 N, 40 N<br/>b. 75 N, 25 N<br/>c. 65 N, 35 N<br/>d. 55 N, 45 N</p>Solution :<img src="https://d10lpgp6xz60nq.cloudfront.net/physics_images/NCERT_OBJ_FING_PHY_XI_C07_E01_039_S01.png" width="80%"/> <br/> AB is the rod. `K_(1)` and `K_(2)` are the two knife <a href="https://interviewquestions.tuteehub.com/tag/edges-965708" style="font-weight:bold;" target="_blank" title="Click to know more about EDGES">EDGES</a>. Since the rod is uniform, therefore its weight acts at its center of <a href="https://interviewquestions.tuteehub.com/tag/gravity-18707" style="font-weight:bold;" target="_blank" title="Click to know more about GRAVITY">GRAVITY</a> G. <br/> Let `R_(1)` and `R_(2)` be reactions at the knife edges. For the translational equilibrium of the rod, <br/> `R_(1)+R_(2)-60N-40N=0` <br/> `R_(1)+R_(2)=60N+40N=100N` <br/> For the rotational equilibrium, talking moments about G, we <a href="https://interviewquestions.tuteehub.com/tag/get-11812" style="font-weight:bold;" target="_blank" title="Click to know more about GET">GET</a> <br/> `-R_(1)(40)+60(2)+R_(2)(40)=0` <br/> `R_(1)-R_(2)=1200/40=30 N`..............(ii) <br/> Adding (i) and (ii), we get `2R_(1)=130N` or `R_(1)=65N` Substituting this <a href="https://interviewquestions.tuteehub.com/tag/value-1442530" style="font-weight:bold;" target="_blank" title="Click to know more about VALUE">VALUE</a> in Eq. (i), we get `R_(2)=35N`</body></html>


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