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A block of mass m is placed on a rough inclined plane. The coefficient of friction between the block and the plane is mu and the inclination of the plane is theta. Initially theat-0 and the block will certain stationary on the plane. Now the inclination theta is gradually increased. The block presses the inclined plane with a force mg cos theta. So welding strength between the block and inclined is mu mgcos theta, and the pulling forces is mg sin theta. As soon as the pulling force is greater than the welding strength, the welding breaks and the block starts sliding, the angle theta for which the block starts sliding is called angle of repose (lambda). During the contact, two contact forces are acting between the block and the inclined plane. The pressing reaction (Normal reaction) and the shear reaction ( frictional force) and the shear reaction(frictional force) . The net contact force will be resultant of both. If the entire system, were accelerated upward with acceleration 'a', the angle of repose , would : |
Answer» <html><body><p>increases<br/>decreases<br/>remain same <br/>increase of `agtg`</p><a href="https://interviewquestions.tuteehub.com/tag/solution-25781" style="font-weight:bold;" target="_blank" title="Click to know more about SOLUTION">SOLUTION</a> :Angle `(theta')` fo repose `:` <br/> `m(g+a)sintheta'=F` <br/> `m(g+a)<a href="https://interviewquestions.tuteehub.com/tag/cos-935872" style="font-weight:bold;" target="_blank" title="Click to know more about COS">COS</a> theta'=R` <br/> `:. (F)/(R)=<a href="https://interviewquestions.tuteehub.com/tag/tantheta-3098264" style="font-weight:bold;" target="_blank" title="Click to know more about TANTHETA">TANTHETA</a>'` <br/> `theta'=tan^(-1)((F)/(R))=alpha` <br/> Hence angle of repose does not change.</body></html> | |