Question

a 60m ladder weighing 100N rests against a wall at a point 48 degree north above the ground. the center of gravity of the ladder is 1/3 the way up. a 160N man climbs halfway up the ladder assuming that the wall is friction less. find the reaction on the wall and the frictional force.

Answer 1

Question #60869, Physics / Classical Mechanics | for completion

a 60m ladder weighing 100N rests against a wall at a point 48 degree north above the ground. the center of gravity of the ladder is 1/3 the way up. a 160N man climbs halfway up the ladder assuming that the wall is friction less. find the reaction on the wall and the frictional force.

$$\begin{array}{l} L = 6 0 m \\ F _ {1} = 1 0 0 N \\ \alpha = 4 8 ^ {0} \\ F _ {m} = 1 6 0 N \\ N _ {a} -? F _ {f r} -? \\ \end{array}$$

Solution:

On the basis of the vanishing of the sum of the moments of all forces about the axis passing through the point B, we form the equation:

$$\begin{array}{l} N _ {A} L \sin \alpha - F _ {m} \frac {L}{2} \cos \alpha - F _ {f} \frac {L}{3} \cos \alpha = 0 \\ N _ {A} L \sin \alpha = L \cos \alpha \left(\frac {F _ {m}}{2} - \frac {F _ {l}}{3}\right) \\ N _ {A} = t g \alpha \left(\frac {F _ {m}}{2} - \frac {F _ {l}}{3}\right) \\ N _ {A} = t g 4 8 ^ {0} \left(\frac {1 6 0 N}{2} - \frac {1 0 0 N}{3}\right) \approx 5 1, 8 3 N - \text {the reaction force of the wall} \\ \end{array}$$

As the ladder is fixed, then the equality of forces:

$N_{A} = F_{fr} = 51,83N$ - the frictional force

Answer the questions:

$N_{A} = 51,83N$

$N_{A} = F_{fr} = 51,83N$

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