A 30-kg uniform ladder 12 m long is leaning against a frictionless wall at an angle of 50 degrees above the horizontal. A 45-kg boy climbs 5.0 m up the ladder. What is the magnitude of the force exerted by the ladder on the wall?
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To find the magnitude of the force exerted by the ladder on the wall, we need to analyze the forces acting on the ladder and set up an equation of equilibrium.
First, let's analyze the forces acting on the ladder:
1. Weight of the ladder (Wl): The weight of the ladder is equal to the mass of the ladder (30 kg) multiplied by the acceleration due to gravity (9.8 m/s^2). So, Wl = 30 kg * 9.8 m/s^2 = 294 N. The weight acts vertically downward.
2. Weight of the boy (Wb): The weight of the boy is equal to the mass of the boy (45 kg) multiplied by the acceleration due to gravity (9.8 m/s^2). So, Wb = 45 kg * 9.8 m/s^2 = 441 N. The weight acts vertically downward.
3. Normal force exerted by the wall (Nw): The normal force exerted by the wall acts perpendicular to the wall and opposes the weight of the ladder and the boy.
4. Force exerted by the boy (Fb): The force exerted by the boy acts vertically upward and is equal to the weight of the boy (Wb = 441 N).
Now, let's set up the equation of equilibrium in the vertical direction:
ΣFy = 0
The upward forces are positive, and the downward forces are negative:
+Fb - Wl - Wb - Nw = 0
Substituting the values, we get:
441 N - 294 N - 441 N - Nw = 0
Simplifying the equation:
-294 N - Nw = 0
Nw = -294 N (Since Nw is acting downward)
Now, the magnitude of the force exerted by the ladder on the wall is the absolute value of Nw:
Magnitude of Nw = |Nw| = |-294 N| = 294 N
Therefore, the magnitude of the force exerted by the ladder on the wall is 294 N.