In Fiz a box of am aunts (total mass m_1 = 1.6 kg) and a box of ant uncles (tota

Question

In Fiz a box of am aunts (total mass m_1 = 1.6 kg) and a box of ant uncles (total mass m_2 = 3.30 kg) slide down an inclined plane while attached by a massless rod parallel to the plane. The angle of incline is theta = 30.0 degree. The coefficient of kinetic friction between the aunt box and the incline is mu_1 = 0.22 ; that between the uncle box and the incline is mu_2= 0.11. Compute the tension in the rod and the magnitude of the common acceleration of the two boxes, How would the answers to (a) and (b) change if the uncles trailed the aunts?

Explanation / Answer

Gravitational Forces on the boxes parallel to the motion are determined by

Fg = mg sin where is the angle of the incline.
The Fg aunts = (1.6)(9.81)sin 30° = 7.85 N
The Fg uncles = (3.3)(9.81)sin 30° = 16.2 N
For a total of 24.05 N

G force perpendicular to the plane = mg cos and are
(1.6)(9.81)cos 30°= 13.6 N and
(3.3)(9.81)cos 30°= 28.04 N, respectively.
The Normal forces have equal magnitude and the frictional forces are calculated by mulitpying by
Aunts Friction = 13.6 N(0.22) = 2.99 N
Uncles Friction = 28.04 N (0.11) = 3.08 N
For a total of 6.07 N.

The gravity and friction are working in opposite directions, to the total force on the system is 24.05 N - 6.07 N= 17.98 N
Acceleration is F/m = 17.98 N/(1.6 + 3.3)kg = 3.7 m/s^2

a) and b) Now we go back to the aunts. We know mass and acceleration, so the net force must be

(1.6 kg)(3.7 m/s^2)= 5.92 N
But gravity and friction account for 7.85 - 2.99 = 4.86 N
Tension on the rod must provide a (5.92 - 4.86) = 1.06 N force on the ant aunt's down the incline.

c) Checking in on the ant uncles, the net force must be:
(3.3 kg)(3.7 m/s^2) = 12.21 N
Gravity and tension combine for 16.2 - 3.08 = 13.12 N
So the rod must slow the uncles down by 0.91 N

Tension on the rod = 1.06 N. It pulls up the incline on the uncles and down the incline on the aunts.

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