MTH304 Assignment 1 Solution and Discussion


  • Cyberian's Gold

    Assignment # 1 MTH304 (Fall 2019)

    Maximum Marks: 20
    Due Date: 17 -11-2019

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    Question No. 1:

    A particle of mass 10kg is placed on an inclined plane which makes and angle of with the horizontal. Find the resolved parts of the weight of the particle in the direction parallel and perpendicular to the plane. MARKS 10

    Question No. 2:

    Two particles of mass 3kg each are connected by a light inextensible string which passes over a smooth fixed pulley, which is attached to a string C. The string C is hanging on the fixed support. The particles are at rest. Find the tension in the string C. MARKS 10


  • Cyberian's Gold

    Solution of Assignment # 1 MTH304 (Fall 2019)

    Maximum Marks: 20
    Due Date: 17 -11-2019

    Question No. 1:

    A particle of mass 10kg is placed on an inclined plane which makes and angle of with the horizontal. Find the resolved parts of the weight of the particle in the direction parallel and perpendicular to the plane. MARKS 10
    Solution
    3480151d-364b-4d32-8c7d-08ecd937e68c-image.png

    Question No. 2:

    Two particles of mass 3kg each are connected by a light inextensible string which passes over a smooth fixed pulley, which is attached to a string C. The string C is hanging on the fixed support. The particles are at rest. Find the tension in the string C. MARKS 10

    Solution
    f2e46de8-2a58-4ad4-a49e-eb3c478212e0-image.png


  • Cyberian's Gold

    Solution of Assignment # 1 MTH304 (Fall 2019)

    Maximum Marks: 20
    Due Date: 17 -11-2019

    Question No. 1:

    A particle of mass 10kg is placed on an inclined plane which makes and angle of with the horizontal. Find the resolved parts of the weight of the particle in the direction parallel and perpendicular to the plane. MARKS 10
    Solution
    3480151d-364b-4d32-8c7d-08ecd937e68c-image.png

    Question No. 2:

    Two particles of mass 3kg each are connected by a light inextensible string which passes over a smooth fixed pulley, which is attached to a string C. The string C is hanging on the fixed support. The particles are at rest. Find the tension in the string C. MARKS 10

    Solution
    f2e46de8-2a58-4ad4-a49e-eb3c478212e0-image.png


  • Cyberian's Gold

    @zareen said in MTH304 Assignment 1 Solution and Discussion:

    Two particles of mass 3kg each are connected by a light inextensible string which passes over a smooth fixed pulley, which is attached to a string C. The string C is hanging on the fixed support. The particles are at rest. Find the tension in the string C.

    2407d6a7-e40d-43e1-aa9b-88642cab4d66-image.png

    Replace 6 with 3


  • Cyberian's Gold

    @zareen said in MTH304 Assignment 1 Solution and Discussion:

    A particle of mass 10kg is placed on an inclined plane which makes and angle of with the horizontal. Find the resolved parts of the weight of the particle in the direction parallel and perpendicular to the plane. MARKS 10

    81a39245-156d-4127-a5a8-74a6f246e856-image.png

    Free Body Diagram
    T tension of string, W weight of the box, N force normal to and exerted by the inclined plane on the box, Fs is the force of friction

    Forces and their components on the x-y system of axis.
    81fe9931-db59-42f5-a7ce-938e402716eb-image.png

    cbb4c798-ca62-479a-829c-dc751b462cd0-image.png

    Equilibrium: W + T + N + Fs = 0
    Forces represented by their components
    W = (Wx , Wy) = ( - M g sin(35°) , - M g cos(35°))
    T = (Tx , Ty) = (|T| cos (25°) , |T| sin (25°) )
    N = (0 , Ny) = (0 , |N|)
    Fs = (- |Fs| , 0) = ( - μs |N| , 0) , where μs is the coefficient of friction between the box and the inclined plane.
    Sum of x components = 0

    • M g sin(35°) + |T| cos (25°) + 0 - μs |N| = 0
      which may be rewritten as
      |T| cos (25°) = μs |N| + M g sin(35°)
      sum of y components = 0
      • M g cos(35°) + |T| sin (25°) + |N| + 0 = 0
        |T| sin (25°) = M g cos(35°) - |N|
        We now need to solve the system of two equations with two unknowns |T| and |N|.
        |T| cos (25°) = μs |N| + M g sin(35°) (equation 1)
        |T| sin (25°) = M g cos(35°) - |N| (equation 2)
        solve equation 2 above for |N| to get
        |N| = M g cos(35°) - |T| sin (25°)
        Substitute |N| by M g cos(35°) - |T| sin (25°) in eq 1 to get
        |T| cos (25°) = μs [ M g cos(35°) - |T| sin (25°) ] + M g sin(35°)
        rewrite above equation as follows
        |T| [ cos (25°) + μs sin (25°) ] = μs M g cos(35°) + M g sin(35°)
        Solve for |T|
        |T| =

    μs M g cos(35°) + M g sin(35°)
    cos (25°) + μs sin (25°)
    Substitute with numerical Values
    μs = 0.3, M = 10 Kg, g = 10 m/s^2
    |T| ≈ 79.3 N

    Use |N| = M g cos(35°) - |T| sin (25°) found above
    |N| = 100 cos(35°) - 79.3 sin (25°) ≈ 48.4 N



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