Girman karfin wutar lantarki - matsaloli da mafita

1. Ana amfani da ƙarfin P a ƙarshen wani katako mai tsawon mita 2. Menene girman ƙarfin juyi ? Axis na juyawa a wurin A.

An sani:

Ƙarfi (F) = 10 N

Tsawon AB (r _AB ) = 2 m

Ƙarfin F yana tsaye a kan katakon.

Hannun lever (l ) = r _AB sin 90 ^o = (2 m)(1) = 2 m

Ana so: Ƙarfin juyi game da axis na juyawa

Magani:

Ƙarfin juyi:

τ = F l = (10 N)(2 m) = 20 N m

Alamar ƙarin saboda hasken yana juyawa a akasin agogo.

[irp]

2. Tsawon katakon AB shine mita 2 kuma girman ƙarfin F shine N10. Menene girman karfin juyi? Axis na juyawa a wurin A.

An sani:

Ƙarfi (F) = 10 N

Tsawon AB (r _AB ) = 2 m

Hannun lever (l ) = r _AB sin 60 ^o = (2 m)(0.5 √3 ) = √3 m

Ana so: Ƙarfin juyi game da axis na juyawa

Magani:

Ƙarfin juyi:

τ = F l = (10 N)( √3 m) = 10 √3 N m

Alamar ƙari saboda ƙarfin F yana sa hasken ya juya juyawar agogon da ke juyawa.

[irp]

3. Tsawon katako mita 2 ne. Girman F ₁ shine 10 N kuma girman F ₂ shine 15 N. Kayyade karfin wutar lantarki da ke kewaye da tsakiyar katakon.

Axis na juyawa a tsakiyar hasken.

An sani:

Ƙarfi 1 (F ₁ ) = 10 N

Nisa tsakanin F ₁ da tsakiyar katako (r ₁ ) = 1 m

Hannun lever 1 (l ₁ ) = r ₁ sin 90 ^o = (1 m)(1) = 1 m

Ƙarfin F ₁ yana tsaye a kan katako.

Ƙarfi 2 (F ₂ ) = 15 N

Nisa tsakanin F ₂ da tsakiyar katakon (r ₂ ) = 1 m

Ƙarfin F ₂ yana tsaye a kan katako.

Hannun lever 2 (l ₂ ) = r ₁ sin 90 ^o = (1 m)(1) = 1 m

Ana so: Ƙarfin juyi mai ƙarfi a kusa da axis na juyawa

Magani:

Ƙarfin juyi 1:

τ ₁ = F ₁ l ₁ = (10 N)( 1 m) = 10 N m

Alamar ƙari saboda ƙarfin F ₁ yana sa hasken ya juya juyawar agogon da ke gaba da agogo.

Ƙarfin juyi 2:

τ ₂ = F ₂ l ₂ = (15 N) ( 1 m) = -15 N

Alamar ragewa saboda ƙarfin F ₂ yana sa hasken ya juya a hannun agogo.

Ƙarfin wutar lantarki:

Σ τ = τ ₁ - τ ₂ = 10 - 15 = - 5 N m

Alamar ragewa ita ce saboda hasken yana juyawa a hannun agogo.

[irp]

4. Tsawon katakon AB shine mita 2, girman F ₁ shine 10 N kuma girman F ₂ shine N 10. Kayyade karfin wutar lantarki da ke kewaye da tsakiyar katakon.

Axis na juyawa a tsakiyar hasken.

An sani:

Ƙarfi 1 (F1 ₎ = 10 Nn

Nisa tsakanin F ₁ da tsakiyar katakon (r ₁ ) = 1 m

Hannun lever 1 (l ₁ ) = r ₁ sin 60 ^o = (1 m)(0.5 √3 ) = 0.5 √3 m

Ƙarfi 2 (F ₂ ) = 10 N

Nisa tsakanin F ₂ da tsakiyar katakon (r ₂ ) = 1 m

Ƙarfin F ₂ yana tsaye a kan katako.

Hannun lever 2 (l ₂ ) = r ₂ sin 90 ^o = (1 m)(1) = 1 m

Ana so: Ƙarfin juyi mai ƙarfi game da axis na juyawa

Magani:

Ƙarfin juyi 1:

τ ₁ = F ₁ l ₁ = (10 N) ( 0.5 √3 m) = 5 √3 = 8.7 Nm

Alamar ƙari saboda ƙarfin F ₁ yana sa hasken ya juya a hannun agogo.

Ƙarfin juyi 2:

τ ₂ = F ₂ l ₂ = (10 N)( 1 m) = -10 N m

Alamar ragewa saboda ƙarfin F ₂ yana sa hasken ya juya a hannun agogo.

Ƙarfin wutar lantarki:

Σ τ = τ ₁ - τ ₂ = 8.7 - 10 = - 1.3 N m

Alamar ragewa saboda ƙarfin yanar gizo yana sa hasken ya juya a hannun agogo.

[irp]

5. Tsawon katako shine 10 m, girman F ₁ shine 10 N, girman F ₂ shine 10 N kuma girman F ₃ shine 15 N. Nisa tsakanin maki A da maki C shine 7.5 m. Ƙarfin F ₂ yana tsakiyar katako. Ka ƙayyade ƙarfin wutar lantarki game da maki C wanda yake a mita 2.5 daga maki B.

Axis na juyawa yana nan a wurin C

An sani:

Ƙarfi 1 (F ₁ ) = 10 N

Nisa tsakanin F ₁ da maki C (r ₁ ) = 2.5 m

Ƙarfin F ₁ yana tsaye a kan katako.

Hannun lever 1 (l ₁ ) = r ₁ sin 90 ^o = ( 2.5 m)( 1 ) = 2.5 m

Ƙarfi 2 (F ₂ ) = 10 N

Nisa tsakanin F ₂ da maki C (r ₂ ) = 2.5 m

Ƙarfin F ₂ yana tsaye a kan katako.

Hannun lever 2 (l ₂ ) = r ₂ sin 90 ^o = ( 2.5 m)(1) = 2.5 m

Ƙarfi 3 (F ₃ ) = 15 N

Nisa tsakanin F ₃ da maki C (r ₃ ) = 7.5 m

Ƙarfin F ₃ yana tsaye a kan katako.

Hannun lever 3 (l ₃ ) = r ₃ sin 90 ^o = (7.5 m)(1) = 7.5 m

Ana so: Ƙarfin juyi mai ƙarfi game da axis na juyawa

Magani:

Ƙarfin juyi 1:

τ ₁ = F ₁ l ₁ = (10 N)( 2.5 m) = 25 N m

Alamar ƙari saboda ƙarfin F ₁ yana sa hasken ya juya a hannun agogo.

Ƙarfin juyi 2:

τ ₂ = F ₂ l ₂ = (10 N)( 2.5 m) = 25 N m

Alamar T da ƙari saboda ƙarfin F ₂ yana sa hasken ya juya a hannun agogo.

Ƙarfin juyi 3:

τ ₃ = F ₃ l ₃ = (15 N)( 7.5 m) = -112..5 N m

Alamar ragewa saboda ƙarfin F ₃ yana sa hasken ya juya a hannun agogo.

Ƙarfin wutar lantarki:

Σ τ = τ ₁ + τ ₂ – τ ₃ = 25 + 25 – 112.5 = – 62.5 N m

Alamar ragewa saboda ƙarfin yanar gizo yana sa hasken ya juya a hannun agogo.

[irp]

6. Tsawon katako mita 10 ne, girmansa kuma mita F ne₁ shine 10 N, girman F₂ shine 10 N kuma girman F₃ shine 10 N. Ƙayyade ƙarfin wutar lantarki game da wurin A, wanda yake da nisan mita 5 daga poiamfani da ƙarfi F₁.

Axis na juyawa a wurin A.

An sani:

Ƙarfi 1 (F ₁ ) = 10 N

Nisa tsakanin F ₁ da maki A (r ₁ ) = 5 m

Hannun lever 1 (l ₁ ) = r ₁ sin 60 ^o = (5 m)(0.5 √3 ) = 2.5 √3 m

Ƙarfi 2 (F ₂ ) = 10 N

Nisa tsakanin F ₂ da maki A (r ₂ ) = 0

Ƙarfin F ₂ yana tsaye a kan katako.

Hannun lever 2 (l ₂ ) = r ₂ zunubi 90 ^o = (0)(1) = 0

Ƙarfin 3 (F3 ₎ = 10 N

Nisa tsakanin F ₃ da maki A (r ₃ ) = 10 m

Hannun lever 3 (l ₃ ) = r ₃ sin 30 ^o = (10 m)(0.5) = 5 m

Ana so: karfin juyi mai ƙarfi game da axis na juyawa

Magani:

Ƙarfin juyi 1:

τ ₁ = F ₁ l ₁ = (10 N)( 2.5 √3 m ) = 25√3 = 43.3 N m

Alamar ƙari saboda ƙarfin F ₁ yana sa hasken ya juya a hannun agogo.

Ƙarfin juyi 2:

τ ₂ = F ₂ l ₂ = (10 N) (0) = 0

Ƙarfin juyi 3:

τ ₃ = F ₃ l ₃ = (10 N)( 5 m) = -50 N m

Alamar ragewa saboda ƙarfin F ₃ yana sa hasken ya juya a hannun agogo.

Ƙarfin wutar lantarki:

Σ τ = τ ₁ + τ ₂ – τ ₃ = 43.3 + 0 – 50 = – 6.7 N m

Alamar ragewa saboda ƙarfin yanar gizo yana sa hasken ya juya a hannun agogo.