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

Question

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?

Answer 1

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:

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

The forces of friction are

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

Substitute them to find the mass of the box A:

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

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

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