Kev sib npaug ntawm lub zog ib txwm muaj

3 cov lus nug txog kev sib npaug ntawm lub zog ib txwm muaj

1. Ib lub thaiv muaj qhov hnyav ntawm 5 kg. Yog tias g = 10 m/s², txiav txim siab:

a) qhov hnyav ntawm lub thaiv

b) lub zog ib txwm yog tias lub thaiv raug muab tso rau ntawm lub dav hlau tiaj tus

c) lub zog ib txwm yog tias lub thaiv nyob rau ntawm lub dav hlau uas tsim lub kaum sab xis ntawm 30^o mus rau kab rov tav

Paub:

Qhov hnyav ntawm lub thaiv (m) = 5 kg

Kev nrawm vim lub ntiajteb txawj nqus (g) = 10 m/s²

SE pob: w, N ntawm lub dav hlau thiab N ntawm lub toj

tshuaj:

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?

Paub Tias:

Qhov hnyav (m) = 120 grams = 0.12 kg

Lub zog (F) = 15 Newton

SE pob: ib txwm muaj zog

tshuaj:

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.

Paub:

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

SE pob: Normal force on cardboard (N)

tshuaj:

Calculate the normal force using the normal force formula:

N = w – F_y = 200 – 120 = 80 Newtons