Ekwazzjoni tal-forza normali

3 mistoqsijiet dwar l-ekwazzjoni tal-forza normali

1. Blokk għandu massa ta' 5 kg. Jekk g = 10 m/s², iddetermina:

a) il-piż tal-blokka

b) il-forza normali jekk il-blokka titqiegħed fuq pjan ċatt

c) il-forza normali jekk il-blokka tistrieħ fuq pjan inklinat li jifforma angolu ta' 30^o għall-orizzontali

Magħruf:

Il-massa tal-blokka (m) = 5 kg

Aċċelerazzjoni minħabba l-gravità (g) = 10 m/s²

Mfittxija: w, N fuq il-pjan u N fuq l-inklinazzjoni

soluzzjoni:

a) Beam weight

w = m g = 5 (10) = 50 Newton

b) The normal force if the block is in a flat plane

N = w = 50 Newton

c) The normal force if the block is on an inclined plane

N = w_y = w cos θ = 50 (cos 30) = 50 (½ √3) = 25√3 Newton

2. A 120 gram eraser is pressed perpendicularly to the board with a force of 15 N. What is the normal force acting on the eraser?

Magħruf:

Massa (m) = 120 gramma = 0.12 kg

Forza (F) = 15 Newton

Mfittxija: Forza normali

soluzzjoni:

Normal force = thrust = 15 Newton

3. Bottled drinking water that is still in the box is pulled with a force of F = 200 N which forms an angle of 37^o with the horizontal. Sin 37^o = 0.6. The mass of the cardboard and its contents is 20 kg. If the coefficient of static friction on the floor is 0.5 and the coefficient of kinetic friction on the floor is 0.2, determine the normal force acting on the cardboard.

Magħruf:

Pull force (F) = 200 N

Sin 37 = 0,6

Fy = F sin 37 = (200)(0,6) = 120 N

Cardboard mass (m) = 20 kg

The weight of the cardboard (w) = m g = (20)(10) = 200 N

Mfittxija: Normal force on cardboard (N)

soluzzjoni:

Calculate the normal force using the normal force formula:

N = w – F_y = 200 – 120 = 80 Newton