A student obtained a sharp image of the grill of a window on a screen, using a convex lens. For getting better results, the teacher suggested focusing on a distant tree instead of the grill. In which the direction should the lens be moved for this purpose?

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Explanation:

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The magnitude of the forces acting at the top are;

= 132.95 N

= 0

The magnitude of the forces acting at the bottom are;

= = -132.95 N

= 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;

= 0

Therefore;

=

Where;

= The sum of clockwise moments about B

= The sum of counterclockwise moments about B

Therefore, we have;

= 2  × (2/6) × 10.0 × 9.81 + 3.0 × (2/6) × 70 × 9.81

= × √(6² - 2²)

Therefore, we get;

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

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

The reaction force on the wall, ≈ 132.95 N

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

Therefore;

=

= - ≈ -132.95 N

Similarly, at equilibrium, we have;

∑Fₓ = = 0

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

= -70.0 × 9.81 - 10 × 9.81 +

∴ The upward force acting at the bottom, = 784.8 N

Therefore;

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

At the top;

= ≈ 132.95 N←

= 0 (The surface upon which the ladder rest at the top is frictionless)

At the bottom;

= ≈ -132.95 N →

= = 784.8 N ↑

Learn more about equilibrium of forces here;