Question

A uniform ladder stands on a rough floor and rests against a frictionless wall. Since the floor is rough, it exerts both a normal force N1 and a frictional force f1 on the ladder. However, since the wall is frictionless, it exerts only a normal force N2 on the ladder. The ladder has a length of L = 4.6m, a weight of WL= 69.0N , and rests against the wall a distance d = 3.75 m above the floor. If a person with a mass of m = 90 kg is standing on the ladder, determine the forces exerted on the ladder when the person is halfway up the ladder.
Required:
Solve of N1, N2 and f1

Answer 1

Answer:

The  normal force N1 exerted by the floor is  $N_1 = 951 \ N$

The  normal force N2 exerted by the wall is  $N_2= 616.43 \ N$

The frictional force exerted by the wall is  $f = N_2 = 616.43 \ N$  

Explanation:

From the question we are told that

    The length of the ladder is  $L = 4.6 \ m$

    The weight of the ladder  is

    The distance of the ladder position on the wall from the floor is  $D = 3.75 \ m$

     The mass of the person is  $m = 90 kg$

Applying Pythagoras theorem

The length of the position the ladder on the ground from the base of the wall is

    $A = \sqrt{L^ 2 - D^2}$

substituting values

    $A = \sqrt{(4.6^2)-(3.75^2)}$

    $A = 2.66 \ m$

  In order the for the ladder not to shift from the ground the sum of the moment about the position of the ladder on the ground must be equal to zero this is mathematically represented as

        $\sum M = 0 = N_2 * D - [\frac{1}{2} * W_L ] * [(mg) *A ]$

         $\sum M = 0 = N_2 * 3.75 - [\frac{1}{2} * 69.0 ] * [(90*9.8) * \frac{4.6}{2.66} ]$

        $N_2 * 3.75 =2311.62$

        $N_2 * 3.75 =2311.62$

        $N_2= 616.43 \ N$

Now the force exerted by the floor on the ladder is mathematically represented as

           $N_1 = W_L + (m * g )$

substituting values

          $N_1 = 951 \ N$

Now the horizontal forces acting on the ladder are $N_2 \ and \ f$ and they are in opposite direction so

     $f = N_2 = 616.43 \ N$