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I. (25 points) T3 θ m1 T1 m2 T2 m3 This is the same situation as problem I on Exam 1, except now there's friction!!! Three boxes (with masses $m_1$, $m_2$, and $m_3$, as shown in the diagram) are on a flat surface with coefficient of kinetic friction $\mu_k$, and are connected with two massless ropes. You pull on a third rope, giving it a force $T_3$ at an angle $\theta$ above the horizontal, and the three boxes move to the right (all with the same acceleration). The tensions in the two ropes connecting the boxes are $T_1$ and $T_2$. 1.1 (9 pts) Draw three separate free-body diagrams, one for each box. Label all forces. 1.2 (8 pts) Find the magnitude of the total frictional force that the floor exerts on all three boxes. For parts 1.2 express your answers in terms of the variables $m_1$, $m_2$, $m_3$, $T_3$, $\theta$, $g$, and $\mu_k$. (You don't have to use all these variables in each answer, but your answer should include only these variables and no others.) 1.3 (8 pts) Find the acceleration of the boxes (just one answer). (And see above.)

          I. (25 points)
T3
θ
m1  T1  m2  T2  m3
This is the same situation as problem I on Exam 1, except now there's friction!!! Three boxes (with masses $m_1$, $m_2$, and $m_3$, as shown in the diagram) are on a flat surface with coefficient of kinetic friction $\mu_k$, and are connected with two massless ropes. You pull on a third rope, giving it a force $T_3$ at an angle $\theta$ above the horizontal, and the three boxes move to the right (all with the same acceleration). The tensions in the two ropes connecting the boxes are $T_1$ and $T_2$.
1.1 (9 pts) Draw three separate free-body diagrams, one for each box. Label all forces.
1.2 (8 pts) Find the magnitude of the total frictional force that the floor exerts on all three boxes. For parts 1.2 express your answers in terms of the variables $m_1$, $m_2$, $m_3$, $T_3$, $\theta$, $g$, and $\mu_k$. (You don't have to use all these variables in each answer, but your answer should include only these variables and no others.)
1.3 (8 pts) Find the acceleration of the boxes (just one answer). (And see above.)

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i 25 points t3 m1 t1 m2 t2 m3 this is the same situation as problem i on exam 1 except now theres friction three boxes with masses m1 m2 and m3 as shown in the diagram are on a flat surfa 17454

Added by Joshua A.

University Physics with Modern Physics

Hugh D. Young 14th Edition

Instant Answer

Solved by Expert Vishal Gupta

08/06/2022

Step 1

The horizontal component of the force $T_3$ is $T_3 \cos\theta$. The total frictional force $F_f$ is given by $F_f = \mu_k (m_1 + m_2 + m_3)g = \mu_k Mg$. Show more…

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I. (25 points) T3 θ m1 T1 m2 T2 m3 This is the same situation as problem I on Exam 1, except now there's friction!!! Three boxes (with masses $m_1$, $m_2$, and $m_3$, as shown in the diagram) are on a flat surface with coefficient of kinetic friction $\mu_k$, and are connected with two massless ropes. You pull on a third rope, giving it a force $T_3$ at an angle $\theta$ above the horizontal, and the three boxes move to the right (all with the same acceleration). The tensions in the two ropes connecting the boxes are $T_1$ and $T_2$. 1.1 (9 pts) Draw three separate free-body diagrams, one for each box. Label all forces. 1.2 (8 pts) Find the magnitude of the total frictional force that the floor exerts on all three boxes. For parts 1.2 express your answers in terms of the variables $m_1$, $m_2$, $m_3$, $T_3$, $\theta$, $g$, and $\mu_k$. (You don't have to use all these variables in each answer, but your answer should include only these variables and no others.) 1.3 (8 pts) Find the acceleration of the boxes (just one answer). (And see above.)

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00:01 Hi, here in this given problem first of all this is the frictionless table top over it there is a box kept over it and the two boxes are hanging attached with the help of the threads passing over the ideal pulleys put at the edges of this table.

00:31 Here it is like this, another string passing over the pulley like this.

00:42 This is the block a having mass ma, block b, having mass mb, block c having mass mc.

00:53 Its weight, m, c, g acting vertically downward, so tension in it, suppose this is t1.

01:03 Here this will also be t1, weight of the box of mass m .a, m .ag acting vertically down, tension in the string t2, and here also this is t2.

01:19 It is given that ma is more than m c, so this box will be moving down.

01:28 Suppose with an acceleration a, so mb will be moving towards right with the same acceleration a, mc will be moving up with the same acceleration a.

01:39 In the first part of the problem, forces acting on each box are already has been shown in the figure.

01:50 Forces acting on each box has been have been shown in the figure.

02:13 Answer for the first part.

02:14 Then in the second part, we have to obtain an expression for the acceleration in the system for which using free body diagram of ma as it is moving down.

02:38 So net force setting on it will be mag minus t2 and using newton's second law motion, that will be equal to mass, product of mass, with its acceleration...

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