Two boxes connected by a light horizontal rope are on a horizontal surface, as s
ID: 1784642 • Letter: T
Question
Two boxes connected by a light horizontal rope are on a horizontal surface, as shown in the following figure. The coefficient of kinetic friction between each box and the surface is 0.30. One box (box exttip{B}{B}) has mass 4.75 { m kg} , and the other box (box exttip{A}{A}) has mass m . A force F with magnitude 39.3 { m N} and direction 53.1 { m ^circ} above the horizontal is applied to the 4.75 { m kg} box, and both boxes move to the right with an acceleration of exttip{a}{a} = 1.50 { m m/s^2}
1.What is the tension T in the rope that connects the boxes?
2.What is the mass m of the second box? Use the unrounded intermediate answers to calculate the final one.
i just need value and unit
Explanation / Answer
A) B normal force Fn = mg - Fsin = 4.75kg * 9.8m/s² - 39.3N * sin53.1º
Fn = 15.1 N
so its friction force Fb = µ*Fn = 0.3 * 15.1N = 4.53 N
The horizontal (accelerating) component of the applied force is
Fh = 39.3 N * cos53.1º = 23.6 N
The net force horizontally is
Fnet = ma = Fh - Fb - Fa where Fa is A's friction force.
(4.75kg + m) * 1.50m/s² = 23.6N - 4.53N - 0.3 * m * 9.8m/s²
Dropping units for ease (m is in kg):
7.125 + 1.5m = 19.07 - 2.94m
4.44m = 11.945
m = 2.7 kg (B)
tension T = Fnet + Fa = 2.7kg * 1.5m/s² + 0.3 * 2.7kg * 9.8m/s²
T = 11.988 N (A)
Could have found tension without finding the mass of A through the fbd for B:
Fnet = Fh - Fb - T
4.75kg * 1.5m/s² = 23.6N - 4.53N - T
T = 11.945 N
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