Question

A man of weight 667.5N stands on the middle position of a 222.5N ladder. The upper end B rests on the corner of a wall while lower end A is attached to stop to prevent slipping. The height of end B from ground is 3.66m while end A is 1.83m away from wall. Find the reaction forces at A & B.

Answer 1

A man of weight 667.5N stands on the middle position of a 222.5N ladder. The upper end B rests on the corner of a wall while lower end A is attached to stop to prevent slipping. The height of end B from ground is 3.66m while end A is 1.83m away from wall. Find the reaction forces at A & B.

Solution:

$\alpha -$ angle which ladder makes with the ground

$N_{A} -$ reaction force from the ground (end A)

$N_{B} -$ reaction force from the wall (end B)

$P_{1} = 222.5N -$ weight of the ladder

$P_{2} = 667.5N -$ weight of the man

$h = 3.66m -$ height of end B from ground

$d = 1.83m -$ distance from end A to the wall

Angle which ladder makes with the ground:

$$\tan \alpha = \frac {h}{d} = \frac {3 . 6 6 m}{1 . 8 3 m} = 2$$

From the right triangle: $N_{A} = \sqrt{N_{Ax}^{2} + N_{Ay}^{2}}$ (1)

Newton's second law for the ladder (the first law of equilibrium):

$$\vec {P} _ {1} + \vec {P} _ {2} + \vec {N} _ {A} + \vec {N} _ {B} = \vec {0}$$

Projection of the law on the X-axis:

$$x: N _ {B} - N _ {A x} = 0$$

Projection of the law on the X-axis:

$$y: N _ {A y} - P _ {1} - P _ {2} = 0$$ $$N _ {A y} = P _ {1} + P _ {2} = 222.5N + 667.5N = 890N$$

Momentum equation for point A (the second law of equilibrium):

$$A: M _ {P _ {1}} + M _ {P _ {2}} + M _ {N _ {A}} + M _ {N _ {B}} = 0$$

$(M_{N_A} = M_{fr1} = 0, \text{because moment arm of this forces is zero})$

$$M _ {P _ {1}} = - P _ {1} \cdot \frac {d}{\tan \alpha} \text{ (minus sign because of the direction of force)}$$ $$M _ {P _ {2}} = - P _ {2} \cdot \frac {d}{\tan \alpha}$$ $$M _ {N _ {B}} = N _ {B} \cdot h$$ $$\rightarrow (4):$$ $$N _ {B} \cdot h - P _ {2} \cdot \frac {d}{\tan \alpha} - P _ {1} \cdot \frac {d}{\tan \alpha} = 0$$ $$N _ {B} = \frac {d}{h \cdot \tan \alpha} (P _ {1} + P _ {2}) = \frac {1.83 \, m}{3.66 \, m \cdot 2} (222.5N + 667.5N) = 222.5N$$ $$(1): N _ {A} = \sqrt {N _ {A x} ^ {2} + N _ {A y} ^ {2}} = \sqrt {(N _ {B}) ^ {2} + (N _ {A x}) ^ {2}} = \sqrt {(222.5N) ^ {2} + (890N) ^ {2}} = 917.4N$$

Answer: reaction forces at A and B: $N_{A} = 917.4N$, $N_{B} = 222.5N$.