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    A uniform ladder of weight 50.0 N and length 4.00 m is placed against a frictionless wall making an angle of 60.0° with the ground.

    Question
    HLPaper 2

    A uniform ladder of weight 50.0 N and length 4.00 m is placed against a frictionless wall making an angle of 60.0° with the ground.

    1.

    Outline why the normal force acting on the ladder at the point of contact with the wall is equal to the frictional force F between the ladder and the ground.

    [1]
    Verified
    Solution

    «translational equilibrium demands that the»

    resultant force in the horizontal direction must be zero 1 mark

    «hence Nw=FN_w=FNw​=F »

    2.

    Calculate F.

    [2]
    Verified
    Solution

    «clockwise moments = anticlockwise moments»

    50×2cos⁡60=Nw×4sin⁡6050 \times 2 \cos 60 = N_w \times 4 \sin 60 50×2cos60=Nw​×4sin60 1 mark

    « Nw=F=50×2cos⁡604sin⁡60N_w=F=\frac{50 \times 2 \cos 60}{4 \sin 60}Nw​=F=4sin6050×2cos60​ »

    F=14.4F=14.4F=14.4 «N» 1 mark

    3.

    The coefficient of friction between the ladder and the ground is 0.400. Determine whether the ladder will slip.

    [2]
    Verified
    Solution

    maximum friction force = «0.4×500.4 \times 500.4×50 N» = 20 «N» 1 mark

    14.4<2014.4 < 2014.4<20 AND so will not slip 1 mark

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