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June 2016 p41 q5
3930
Two particles of masses 5 kg and 10 kg are connected by a light inextensible string that passes over a fixed smooth pulley. The 5 kg particle is on a rough fixed slope which is at an angle of \(\alpha\) to the horizontal, where \(\tan \alpha = \frac{3}{4}\). The 10 kg particle hangs below the pulley (see diagram). The coefficient of friction between the slope and the 5 kg particle is \(\frac{1}{2}\). The particles are released from rest. Find the acceleration of the particles and the tension in the string.
Solution
First, calculate the normal reaction \(R\) on the 5 kg particle:
\(R = 5g \cos \alpha = 4g\)
The frictional force \(F\) is:
\(F = 0.5 \times 4g = 2g\)
Apply Newton's second law to the 5 kg particle on the slope:
\(T - 2g - 5g \sin \alpha = 5a\)
Since \(\sin \alpha = \frac{3}{5}\), substitute to get: