Answer to Question #125538 in Classical Mechanics for Hussain Ousman Darboe

Question #125538

Two boxes connected by a light horizontal rope are on a horizontal surface. The coefficient of kinetic friction between each box and the surface is 0.30. Box B has a mass of 5kg and box A has a mass of mkg. A force F with a magnitude of 40N and direction of 53.1° above the horizontal is applied on the 5kg box, and both boxes move to the right with an acceleration of 1.5m/s^2.

(A) what is the tension on the rope that connects the two boxes?

(B) what is the mass of the other box?


1
Expert's answer
2020-07-07T10:02:31-0400

Write equations for all forces acting along x- and y-axes according to Newton's second law. Assume that x is positive if the right, y is positive upward:


"Ox: -f_A-f_B+F\\text{ cos}\\theta=(m_A+m_B)a,\\\\\nOy:N_A-m_Ag=0,\\\\\nN_B-m_Bg+F\\text{ sin}\\theta=0."

The forces of friction are


"f_A=\\mu N_A=\\mu m_Ag,\\\\\nf_B=\\mu N_B=\\mu( m_Bg-F\\text{ sin}\\theta)."

Substitute them to find the mass of the box A:


"F\\text{ cos}\\theta-\\mu [m_Ag+m_Bg-F\\text{ sin}\\theta]=\\\\=(m_A+m_B)a,\\\\\\space\\\\\nm_A=\\frac{F(\\text{cos}\\theta+\\mu\\text{ sin}\\theta)}{\\mu g+a}-m_B\n\\\\\\space\\\\\nm_A=\\frac{40(\\text{cos}53.1\u00b0+0.3\\text{ sin}53.1\u00b0)}{0.3\\cdot9.8+1.5}-5=2.57\\text{ kg}."


The tension in the rope can be found if we apply Newton's second law for box A:


"A:Ox:T-\\mu m_Ag=m_aa,\\\\\nT=m_A(\\mu g+a)=11.4\\text{ N}."

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