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A 60kg60 kg board that is 6 m6 m long is placed at the edge of a platform, with 4 m4 m of its length extending over the edge. The board is held in place by blocks of masses M1M1 and M2M2 placed with their centers of mass on either end. If M2=30kgM2=30kg , what is the minimum value of M1M1 needed to keep the board from falling off the platform? 30kg

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okpalawalter8

Complete Question

The complete question is shown on the first and second uploaded image

Answer:

The minimum mass of [tex]M_1 = 90\ kg[/tex] correct option is  E

Explanation:

 Free body diagram of the set up  in the question is shown on the third uploaded image

  The mass of board is  [tex]M = 60kg[/tex]

   The length of the board is [tex]L = 6 \ m[/tex]

    The length extending over the edge is [tex]L_e = 4 \ m[/tex]

    The second mass is  [tex]M_2 = 30kg[/tex]

Now to obtain [tex]M_1[/tex] we take moment about the edge of the platform

               [tex]M_1 g L_1 = Mg \frac{L}{2} + M_2 g L_2[/tex]

              [tex]M_1 L_1 = M \frac{L}{2} + M_2 L_2[/tex]

  Substituting value  

               [tex]M_1 (2) = (60)(1) + (30)(4)[/tex]

               [tex]M_1 = 90 \ kg[/tex]

Ver imagen okpalawalter8
Ver imagen okpalawalter8
Ver imagen okpalawalter8
ajeigbeibraheem

The minimum value of M1 needed to keep the board from falling off the platform is 90 kg.

From the information given, we are to find:

  • the mass (M1) of placed on the left side edge of the board

Given that:

  • the mass of the board = 60 kg
  • the length of the board = 6 m
  • If the mass on the right side = 30 kg, and the length of the board L1 = 2m
  • Then, the length L2 which extend over the edge = 4m

Consider the center of gravity in the board that lies at the length of the board midpoint.

Then, the distance (D) of the gravity center from the platform end = 3 - 2

= 1 m

∴

Considering the moment about the platform end, the mass (M1) placed on the left side edge of the board can be computed as:

[tex]\mathbf{M_1gL_1 = MgD + M_2gL_2} \\ \\ \mathbf{M_1L_1 = MD + M_2gL_2} \\ \\ \mathbf{M_1(2) = 60 \ kg \times 1 + 30 \ kg \times (4)} \\ \\ \mathbf{ M_1 =\dfrac{60 \ kg + 120 kg }{2} } \\ \\ \mathbf{ M_1 =\dfrac{180 \ kg}{2} } \\ \\ \mathbf{ M_1 =90 \ kg }[/tex]

Therefore, we can conclude that the minimum value M1 needed to keep the board from falling off the platform is 90 kg.

Learn more about masses here:

https://brainly.com/question/19694949?referrer=searchResults

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