A 75 kg window cleaner uses a 10 kg ladder that is 5.0 m long. He places one end on the ground 2.5 m from a wall, rests the upper end against a cracked Window, and climbs the ladder. He is 3.0 m up along the ladder when the window breaks. Neglect friction between the ladder and window and assume that the base of the ladder does not slip. When the window is on the verge of breaking, what are (a) the magnitude of the force on the window from the ladder, (b) the magnitude of the force on the ladder from the ground, and (c) the angle (relative to the horizontal) of that force on the ladder

Solution :

Given : Mass of ladder = 10 kg

Length of ladder = 5 m

Weight of window cleaner = 75 kg

a). Now equate the torque about the lowermost point of the ladder is given by :

Here, N = normal force that the glass exerts on the ladder

Therefore,

                      = 280 N (in 2  significant figures)

b). Equate the forces along horizontal direction,

The horizontal component of the friction,

The vertical component of the friction,

                                                                     = 833 N

Therefore, the net frictional force,

    = 879.7 N

    = 880 N (in 2 significant figures)  

c). The angle the forces makes

  Therefore in 2 significant figures =

adding a(n) to an electric circuit would increase the magnetic field around a wire;

b. solenoid

a solenoid is a type of electromagnet whose purpose is to generate a controlled magnetic field.

if the purpose of the solenoid is instead to impede changes in the electric current, a solenoid can be more specifically classified as an inductor rather than an electromagnet.

i think it would be 5