1. Two boxes are connected by a cord running over a pulley. Ignore the mass of the cord and pulley and any friction in the pulley. Mass of the box 1 = 2 kg, mass of the box 2 = 3 kg, te whakaterenga nā te kaha ā-papa = 10m/s2. Kimihia (a) The acceleration of the system (b) The tension in the cord!
otinga
Mōhiotia:
Taumaha o te pouaka 1 (m1) = 2 kirokaramu
Taumaha o te pouaka 2 (m2) = 3 kirokaramu
Te whakaterenga nā te kaha ā-papa (g) = 10 m/s2
Taumaha o te pouaka 1 (w1) = m1 g = (2)(10) = 20 Newton
Weight of the box 2 (w2) = m2 g = (3)(10) = 30 Newton
Rongoā:
(a) te rahi me te ahunga o te whakaterenga
w2 > w1 no reira te box 2 accelerates downward and the box 1 accelerates upward.
Forces that has the same direction with acceleration (w2 a T1), its sign positive. Forces that has opposite direction with acceleration (T2 a w1), its sign negative.
∑F = ma
w2 - T2 +T1 - w1 = (m1 +m2) a ——-> T1 =T2 =T
w2 – T + T – w1 = (m1 +m2) a
w2 - w1 = (m1 +m2) a
30 – 20 = (2 + 3) he
10 = 5
ā = 10 / 5
a = 2 m/s2
Magnitude of the whakatere he 2 m/s2.
(b) The tension force
The box 2 :
There are two forces acts on the box 2 : first, weight of the box 2 (w2), points downward so it’s positive. Second, tension force exerted on the box 2 (T2), points upward so it’s negative. Apply Te ture tuarua a Newton of motion.
∑F = ma
w2 - T2 = m2 a
30 – T2 = (3)(2)
30 – T2 = 6
T2 = 30 - 6
T2 = 24 Niutona
Box 1 :
There are two forces acts on the box 1. tuatahi, weight of the box 1 (w1), points downward so it’s negative. tuarua, the tension force exerted on the box 1 (T1) points upward so it’s positive. Apply Newton’s second law of motion :
∑F = ma
T1 - w1 = m1 a
T1 – 20 = (2)(2)
T1 - 20 = 4
T1 = 20 + 4
T1 = 24 Niutona
Magnitude of the tension force = T1 =T2 = T = 24 Newton
2. An object on a rough horizontal surface. Mass of the object 1 = 2 kg, mass of the object 2 = 4 kg, acceleration due to gravity = 10 m/s2, coefficient of the static friction = 0.4, coefficient of the kinetic friction = 0.3. The system is at rest or accelerated ? If the system is accelerated, find the magnitude and direction of the system’s acceleration!
otinga
Mōhiotia:
Taumaha o te mea 1 (m1) = 2 kirokaramu
Taumaha o te mea 2 (m2) = 4 kirokaramu
Te whakaterenga nā te kaha ā-papa (g) = 10 m/s2
Coefficient of the te waku pumau (μs) = 0.4
Te tauwehenga o te waku nekeneke (μk) = 0.3
Weight of the object 1 (w1) = m1 g = (2)(10) = 20 Newton
Weight of the object 2 (w2) = m2 g = (4)(10) = 40 Newton
Te kaha noa exerted on the object 1 (N) = w1 = 20 Niutona
Force of the static friction exerted on the object 1 (fs) = μs N = (0.4)(20) = 8 Ngā Niutoni
Force of the kinetic friction exerted on the object 1 (fk) = μk N = (0.3)(20) = 6 Ngā Niutoni
Hiahia: whakaterenga (a)
Rongoā:
w2 > whs (40 Newton > 8 Newton) so the object 2 is accelerated vertically downward and the object 1 is accelerated horizontally rightward. The friction force that acts on the objects 1 is the force of the kinetic friction (fk). Apply Newton’s second law of motion :
∑F = ma
w2 - te = (m1 +m2) a
40 – 6 = (2 + 4) he
34 = 6
a = 34 / 6 = 17 / 3
a = 5.7 m/s2
Te rahi o te whakaterenga = 5.7 m/s2
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- Papatipu me te taumaha
- Te kaha noa
- Te ture tuarua o te nekehanga a Newton
- Te kaha waku
- Te nekehanga i runga i te mata whakapae me te kore he kaha waku
- Te nekehanga o ngā tinana e rua me te tere tere ōrite i runga i te mata whakapae taratara me te kaha waku
- Te nekehanga i runga i te papa whakarara me te kore he kaha waku
- Te nekehanga i runga i te papa whakarara taratara me te kaha waku
- Te nekehanga i roto i te ararewa
- Ka honoa te nekehanga o ngā tinana e ngā taura me ngā pūrere
- E rua ngā tinana he rite te rahi o te whakaterenga
- Te whakaawhiwhi i tētahi piko papatahi – ngā nekehanga porowhita
- Te whakaawhiwhi i tētahi kōpiko peeke – ngā hihiri o te nekehanga porowhita
- Te nekehanga ōrite i roto i te porowhita whakapae
- Te kaha pokapū i roto i te nekehanga porowhita ōrite
