Bir kisi 60 nluk bir kzag 15 derece egimli yoldan tepeye...

Bir kişi 60 N’luk bir kızağı 15 derece eğimli yoldan tepeye doğru sabit hızla çekmektedir. Kişi kızağa bağlanan ipe 25 N’luk kuvvet uyguluyor. İp yatayla 35 derece ’lik açı yapıyorsa, a) Kızakla kar arasındaki kinetik sürtünme katsayısı nedir? b) Kişi tepeye ulaştığında kızağa biner ve aşağıya doğru kaymaya başlarsa aşağı doğru kazandığı ivmenin büyüklüğü ne olur?

Transcript

00:01 Here in this given problem, this is the inclined plane which is inclined at an angle 15 degree.

00:14 A sled is being pulled over it by a man such that the force being applied over it, it is making an angle of 35 degree with the horizontal.

00:38 And as this angle is already we have seen this is 15 degree so this angle remaining here that will be 20 degree so this force is resolved into two components one along the incline upward that is f cos 20 degree another perpendicular to the incline plane that is f sine 20 degree weight of the sled that is given to us as w is equal to 60 newton so this way it is also resolved into two components one of the component perpendicular to the inclined plane that is w cos 15 degree another along the inclined plane downward that is w sine 15 degree.

01:34 Okay.

01:36 In the first part of the problem, there is only one part which we have to solve, that is first part.

01:41 So, as the sled is moving with a uniform motion up, so force of friction, kinetic friction will be acting down.

01:49 This is the normal reaction n exerted by the inclined line plane over the sledge and that normal reaction will be equal to w cos 15 degree minus this component of the applied force f sine 20 degree and force of friction then that will be given by mu k coefficient of kinetic friction mu k times n and as the sledge is moving up with a uniform speed with a constant speed it means there should be no net force acting on it so force acting on it upward which is f cos 20 degree that should be equal to forces acting on it downward and that is w sine 15 degree plus force of friction fk or we can say for the force we know its value.

03:00 This force is given to be 25 newton.

03:06 So this is 25 cos 20 degree weight 60, 60 sin 15 degree plus fk which is mu k.

03:16 Mu k is missing, we have to find it.

03:18 Mu k times n and we know n means normal reaction that is w, w against 60 cos 15 degree minus f force which is 25 sine 20 degree.

03:35 So rearranging the terms we get an expression for this coefficient of friction mu k that will be given by 25 cos 20 degree minus 60 sin 15 degree divided by 60 cos 15 degree minus 25 sin 20 degree...

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