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kvv77 [185]
3 years ago
6

Consider an elevator with a table and a book on top of the table. The mass of the table is 10kg and the mass of the book is 2kg.

Physics
1 answer:
Step2247 [10]3 years ago
4 0

Newton's second law allows us to find the force of the block on the table is 126 N

Newton's second law says that the net force is proportional to the product of the mass and the acceleration of the body

                Σ F = m a

Where the bold letters indicate vectors, m is the mass and the acceleration of the body

A free body diagram is a diagram where the forces are represented without the details of the bodies, in the attached we can see a free body diagram of the system.

Let's start by finding the acceleration of the elevator with kinematics  

                 v = v₀ + a t

                 a = \frac{v-v_o}{a}  

Where v and v₀ are the current and initial velocity, respectively, at acceleration and t is the time

                 a = \frac{8-1}{2}

                 a = 3.5 m / s²

Let's write Newton's second law for each body

The book

                N₂ - W₂ + N₁ = m a

               

Table

                N₁ - W₁ - W₂ = M a

                W₁ = Mg

                W₂ = mg

                N₁ = (M + m) g + M a

                N₁ = (10 + 2) 9.8 + 10 3.5

                N₁ = 152.6 N

This is the reaction of the earth to the support of the block and the table

               N₂ = ma + m g  - N₁

               N₂ = m ( a +g) - N₁  

               N₂ = 2 (3.5 + 9.8) - 152.6

               N₂ = 26.6 - 152.6

               N₂ = -126 N

The negative sign indicates that the direction is opposite to the one assigned, this is the action of the block on the table.

In conclusion using Newton's second law we can find the forces of the block on the table is 126 N

Learn more here: brainly.com/question/19860811

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Explanation:

The free body diagram of the question is shown on the first uploaded image From the question we are told that

           The distance between the two poles is D =14 m

          The mass tied between the two cloth line is  m = 3Kg

         The distance it sags is d_s = 1m

The objective of this solution is to obtain the magnitude of the tension on the ends of the  clothesline

Now the sum of the forces on the y-axis is zero assuming  that the whole system is at equilibrium

       And this can be mathematically represented as

                             \sum F_y = 0

 To obtain \theta we apply SOHCAHTOH Rule

 So    Tan \theta = \frac{opp}{adj}

          \theta = tan^{-1} [\frac{opp}{adj} ]

            = tan^{-1} [\frac{1}{7}]

          =8.130^o

=>  \  \ \ \ \ \ \ \ 2T sin\theta -mg =0

=>  \  \ \ \ \ \ \ \ T =\frac{mg}{2 sin\theta}

=>  \  \ \ \ \ \ \ \ T = \frac{3 * 9.8 }{2 sin \theta }

=>  \  \ \ \ \ \ \ \  T =\frac{29.4}{2sin(8.130)}

=>  \  \ \ \ \ \ \ \  T = 103.96N

             

                 

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