Question

A funicular railway has an angle of incline a equal to 30° to the horizontal (Fig.). A wagon weighing 500 kg-wt descends the incline. Find the tension in the cable during the braking of the wagon at the end of the trip if the speed v₀ of the wagon before slowing down was 2m/sec and the braking time is 5 sec. The coefficient of friction may be taken to be equal to 0.01.

 

Answer 1

Answer: F = 266 kg-wt

Explanation:

The acceleration of the wagon during the braking is a = v₀ /t and is directed upward along the incline. In this direction, there also act on the wagon the force due to the tension in the cable F and the force of friction f = kN = kMg cosα, where N is the normal reaction and M is the mass of the wagon. The component of the weight Mg sinα. is directed downwards along the inclined plane. The equation for Newton's second law for the wagon has the form