Persamaan gaya normal

3 pitakonan babagan persamaan gaya normal

1. Balok duwé massa 5 kg. Yèn g = 10 m/s², nemtokake:

a) bobot blok

b) Gaya normal yen balok dilebokake ing bidang datar

c) Gaya normal yen balok kasebut ana ing bidang miring sing mbentuk sudut 30^o menyang horisontal

Dikenal:

Massa balok (m) = 5 kg

Akselerasi amarga gravitasi (g) = 10 m/s²

Dikarepake: w, N ing bidang lan N ing lereng

Solution:

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?

Dikenal:

Massa (m) = 120 gram = 0.12 kg

Gaya (F) = 15 Newton

Dikarepake: Gaya normal

Solution:

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.

Dikenal:

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

Dikarepake: Normal force on cardboard (N)

Solution:

Calculate the normal force using the normal force formula:

N = w – F_y = 200 – 120 = 80 Newton