Question

To get up on the roof, a person (mass 70.0kg) places a 6.00-m aluminum ladder (mass 10.0 kg) against the house on a concrete pad with the base of the ladder 2.00 m from the house. The ladder rests against a plastic rain gutter, which we can assume to be frictionless. The center of mass of the ladder is 2 m from the bottom. The person is standing 3 meters from the bottom. What are the magnitudes of the forces on the ladder at the top and bottom

Answer 1

The magnitude of the forces acting at the top are;

$\mathbf{F_{Top, \ x}}$ = 132.95 N

$\mathbf{F_{Top, \ y}}$ = 0

The magnitude of the forces acting at the bottom are;

$\mathbf{F_{Bottom, \ x}}$ = $\mathbf{ F_f}$ = -132.95 N

$\mathbf{F_{Bottom, \ y}}$ = 784.8 N

The known parameters in the question are;

The mass of the person, m₁ = 70.0 kg

The length of the ladder, l = 6.00 m

The mass of the ladder, m₂ = 10.0 kg

The distance of the base of the ladder from the house, d = 2.00 m

The point on the roof the ladder rests = A frictionless plastic rain gutter

The location of the center of mass of the ladder, C.M. = 2 m from the bottom of the ladder

The location of the point the person is standing = 3 meters from the bottom

g = The acceleration due to gravity ≈ 9.81 m/s²

The required parameters are;

The magnitudes of the forces on the ladder at the top and bottom

The strategy to be used;

Find the angle of inclination of the ladder, θ

At equilibrium, the sum of the moments about a point is zero

The angle of inclination of the ladder, θ = arccos(2/6) ≈ 70.53 °C

Taking moment about the point of contact of the ladder with the ground, B gives;

$\sum M_B$ = 0

Therefore;

$\sum M_{BCW}$ = $\sum M_{BCCW}$

Where;

$\sum M_{BCW}$ = The sum of clockwise moments about B

$\sum M_{BCCW}$ = The sum of counterclockwise moments about B

Therefore, we have;

$\sum M_{BCW}$ = 2  × (2/6) × 10.0 × 9.81 + 3.0 × (2/6) × 70 × 9.81

$\sum M_{BCCW}$ = $F_R$ × √(6² - 2²)

Therefore, we get;

2  × (2/6) × 10.0 × 9.81 + 3.0 × (2/6) × 70 × 9.81  = $F_R$ × √(6² - 2²)

$F_R$  = (2  × (2/6) × 10.0 × 9.81 + 3.0 × (2/6) × 70 × 9.81)/(√(6² - 2²)) ≈ 132.95

The reaction force on the wall, $F_R$ ≈ 132.95 N

We note that the magnitude of the reaction force at the roof, $F_R$ = The magnitude of the frictional force of bottom of the ladder on the floor, $F_f$ but opposite in direction

Therefore;

$F_R$ = $-F_f$

$F_f$ = - $F_R$ ≈ -132.95 N

Similarly, at equilibrium, we have;

∑Fₓ = $\sum F_y$ = 0

The vertical component of the forces acting on the ladder are, (taking forces acting upward as positive;

$\sum F_y$ = -70.0 × 9.81 - 10 × 9.81 + $F_{By}$

∴ The upward force acting at the bottom, $F_{By}$ = 784.8 N

Therefore;

The magnitudes of the forces at the ladder top and bottom are;

At the top;

$\mathbf{F_{Top, \ x}}$ = $F_R$ ≈ 132.95 N←

$\mathbf{F_{Top, \ y}}$ = 0 (The surface upon which the ladder rest at the top is frictionless)

At the bottom;

$\mathbf{F_{Bottom, \ x}}$ = $F_f$ ≈ -132.95 N →

$\mathbf{F_{Bottom, \ y}}$ = $F_{By}$ = 784.8 N ↑

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