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    Q.1 A box has weight 50N. It is placed on a rough plane inclined 22 degrees to the horizontal, where it remains at rest.
    a) Find the magnitude of the normal reaction force acting on the box.
    b) Find the magnitude of the friction force acting on the box.
    c)The co-efficient of friction between the box and the plane is u. Find an equality that u must satisfy.

    Q.2 The forces F1=(13i-6j)N, F2=(19i+aj)N and F3=(bi+31j)N are in equilibrium. Find a and b.

    Q.3 As a child, of mass 50kg, moves down a slide he travels at a constant speed. The slide is inclined at an angle of 45 degrees to the horizontal. The coefficient of friction between the slide and the child is 0.4
    a) Find the magnitude of the normal reaction acting on the child.
    b) Hence find the magnitude of the friction force acting on the child.
    c) Determine the magnitude of the air resistance force acting on the child.

    Q4. Two ropes are attached to a load of mass 500kg. The ropes make angles of 30 degrees and 45 degrees to the vertical. The tensions in these ropes are T1 and T2 Newtons. The load is also supported by a vertical spring.
    The system is in equilibrium and T1=200.

    a) Show that T2=141, correct to three significant figures.
    b) Find the force that the spring exerts on the load.



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    Bowie88
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    Q.1 A box has weight 50N. It is placed on a rough plane inclined 22 degrees to the horizontal, where it remains at rest.[/B]
    a) Find the magnitude of the normal reaction force acting on the box.
    b) Find the magnitude of the friction force acting on the box.
    c)The co-efficient of friction between the box and the plane is u. Find an equality that u must satisfy.
    a) R = 50cos22 N
    b) F = 50sin22 N
    c) F<=uR => 50sin22<=u50cos22
    u50cos22=>50sin22
    u=>50sin22/50cos22
    u=>tan22

    --------------

    Q.2 The forces F1=(13i-6j)N, F2=(19i+aj)N and F3=(bi+31j)N are in equilibrium. Find a and b.
    Equlibrium
    => F1+F2+F3=0
    => (13i-6j)+(19i+aj)+(bi+31j) = 0

    i-component
    13+19+b = 0
    b = -32

    j-component
    -6+a+31 = 0
    a = 6-31
    a = -25
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    thanks.........
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    any help with the other 2 questions anyone????????
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    (Original post by Bowie88)
    Q.3 As a child, of mass 50kg, moves down a slide he travels at a constant speed. The slide is inclined at an angle of 45 degrees to the horizontal. The coefficient of friction between the slide and the child is 0.4
    a) Find the magnitude of the normal reaction acting on the child.
    b) Hence find the magnitude of the friction force acting on the child.
    c) Determine the magnitude of the air resistance force acting on the child.
    [/B]
    a) Resolve perpendicular:
    R = 50gcos45°
    R = 257.40N

    b) At a constant speed, limiting friction occurs so:
    F = µR
    F = 0.4 x 257.40 = 102.96N

    c) Again, at a constant speed, so resolving parallel there must be no net force:
    Friction + Air Resistance = 50gsin45°
    102.96 + Air Resistance = 257.40
    Air Resistance = 154.44
 
 
 
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