A 0.840- kg glider on a level air track is joined by strings to two hanging mass
ID: 2137763 • Letter: A
Question
A 0.840- kg glider on a level air track is joined by strings to two hanging masses. As seen in the figure, the mass on the left is 4.85 kg and the one on the right is 3.62 kg The strings have negligible mass and pass over light, frictionless pulleys. Find the acceleration of the masses when the air flow is turned off and the coefficient of friction between the glider and the track is 0.35. Take positive to be an acceleration to the right.
Find the tension in the string on the left between the glider and the 4.85- kg mass when the air flow is turned off and the coefficient of friction between the glider and the track is 0.35.
Find the tension in the string on the right between the glider and the 3.62- kg mass when the air flow is turned off and the coefficient of friction between the glider and the track is 0.35.
Explanation / Answer
Since the larger weight is on the left, the system will move to the left. Therefore the glider friction force is to the right.
The forces on the glider are 4.85(9.8) = 47.53 N to the left,
0.84(9.8)(0.43)=3.54 N friction force on the glider to the right,
and 3.62(9.8) = 35.48 N to the right.
Total force from the weights and the friction force = -47.53 + 3.54 + 35.48 = -8.51 N (to the left).
Total mass of the system = 4.85 + 0.84 + 3.62 = 9.31 kg
F=ma or a = F/m = -8.51/9.31 = -0.91 m/s^2 (to the left)
To find tension substitute a in 1 equation:
Fg1-Ft1=m1a
The forces on the glider are 4.85(9.8) = 47.53 N to the left,
0.84(9.8)(0.43)=3.54 N friction force on the glider to the right,
and 3.62(9.8) = 35.48 N to the right.
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