Fysiikan ylioppilaskoe S2008

Ratkaisu

Kirjataan lähtöarvot

m_k=5,2 \text{ kg}, \ m_t = 1,2 \text{ kg}

m ​ k ​ ​ = 5, 2 k g, m ​ t ​ ​ = 1, 2 k g

Piirretään tangon voimakuvio.

Tanko on tasapainossa etenemisen ja pyörimisen suhteen.

\text{NII } \sum \overline{F}=\overline{0}

N I I \sum ​ F ​ ​ ​ = ​ 0 ​ ​ ​

\sum M_A=0

\sum M ​ A ​ ​ = 0

0,42 \text{ m}

0, 42 m

0,65 \text{ m}

0, 65 m

h=0,42 \text{ m}, l=0,65 \text{ m}

h = 0, 42 m, l = 0, 65 m

Ratkaistaan kulma

0,42 \text{ m}

0, 42 m

0,65 \text{ m}

0, 65 m

\tan \color{CornflowerBlue}\alpha=\dfrac{0,42 \text{ m}}{0,65 \text{ m}}

tan α = \frac{​ 0 , 4 2 m ​ ​}{​ 0 , 6 5 m ​}

\color{CornflowerBlue}\alpha

α

\color{CornflowerBlue}\alpha \approx 32,869^\circ

α \approx 32, 86 9 ​ \circ ​ ​

\sum M_A=0

\sum M ​ A ​ ​ = 0

0,42 \text{ m}

0, 42 m

-G_t \dfrac{l}{2} - G_k l +T_y l=0

- G ​ t ​ ​ \frac{​ l ​ ​}{​ 2 ​} - G ​ k ​ ​ l + T ​ y ​ ​ l = 0

T_y l=G_t \dfrac{l}{2} + G_k l

T ​ y ​ ​ l = G ​ t ​ ​ \frac{​ l ​ ​}{​ 2 ​} + G ​ k ​ ​ l

T_y =\dfrac{1}{2}G_t + G_k

T ​ y ​ ​ = \frac{​ 1 ​ ​}{​ 2 ​} G ​ t ​ ​ + G ​ k ​ ​

T \sin \color{CornflowerBlue} \alpha =\dfrac{1}{2}G_t + G_k

T sin α = \frac{​ 1 ​ ​}{​ 2 ​} G ​ t ​ ​ + G ​ k ​ ​

0,65 \text{ m}

0, 65 m

T=\dfrac{\frac{1}{2}G_t + G_k}{\sin \color{CornflowerBlue}\alpha}

T = \frac{​ \frac{​ 1 ​ ​}{​ 2 ​} G ​ t ​ ​ + G ​ k ​ ​ ​ ​}{​ sin α ​}

:\sin \alpha

: sin α

: l

T_y=T \sin \alpha

T ​ y ​ ​ = T sin α

0,42 \text{ m}

0, 42 m

0,65 \text{ m}

0, 65 m

T=\dfrac{\frac{1}{2}G_t + G_k}{\sin \color{CornflowerBlue}\alpha}

T = \frac{​ \frac{​ 1 ​ ​}{​ 2 ​} G ​ t ​ ​ + G ​ k ​ ​ ​ ​}{​ sin α ​}

Lasketaan vaijerin jännitysvoiman suuruus.

T=\dfrac{\frac{1}{2}\cdot 1,2 \text{ kg} \cdot 9,81 \text{ m/s}^2 + 5,2 \text{6 kg} \cdot 9,81 \text{ m/s}^2}{\sin 32,869^\circ}

T = \frac{​ \frac{​ 1 ​ ​}{​ 2 ​} \cdot 1 , 2 k g \cdot 9 , 8 1 m / s ​ 2 ​ ​ + 5 , 2 6 k g \cdot 9 , 8 1 m / s ​ 2 ​ ​ ​ ​}{​ sin 3 2 , 8 6 9 ​ \circ ​ ​ ​}

T \approx 100 \text{ N}

T \approx 100 N

0,42 \text{ m}

0, 42 m

0,65 \text{ m}

0, 65 m

Tehdään voimatarkastelu x- ja y-suunnissa erikseen.

\sum \overline{F}_x=\overline{0}

\sum ​ F ​ ​ ​ ​ x ​ ​ = ​ 0 ​ ​ ​

\overline{N}_x+\overline{T}_x=\overline{0}

​ N ​ ​ ​ ​ x ​ ​ + ​ T ​ ​ ​ ​ x ​ ​ = ​ 0 ​ ​ ​

N_x - T_x =0

N ​ x ​ ​ - T ​ x ​ ​ = 0

N_x =T_x

N ​ x ​ ​ = T ​ x ​ ​

N_x =T \cos \color{CornflowerBlue}\alpha

N ​ x ​ ​ = T cos α

Sovitaan suunnat ylös ja oikealle positiivisiksi.

0,42 \text{ m}

0, 42 m

0,65 \text{ m}

0, 65 m

\sum \overline{F}_y=\overline{0}

\sum ​ F ​ ​ ​ ​ y ​ ​ = ​ 0 ​ ​ ​

\overline{N}_y+\overline{T}_y+\overline{G}_t+\overline{G}_k=\overline{0}

​ N ​ ​ ​ ​ y ​ ​ + ​ T ​ ​ ​ ​ y ​ ​ + ​ G ​ ​ ​ ​ t ​ ​ + ​ G ​ ​ ​ ​ k ​ ​ = ​ 0 ​ ​ ​

N_y +T_y -G_t-G_k =0

N ​ y ​ ​ + T ​ y ​ ​ - G ​ t ​ ​ - G ​ k ​ ​ = 0

N_y=G_t +G_k - T_y

N ​ y ​ ​ = G ​ t ​ ​ + G ​ k ​ ​ - T ​ y ​ ​

N_y=G_t +G_k - T \sin \color{CornflowerBlue}\alpha

N ​ y ​ ​ = G ​ t ​ ​ + G ​ k ​ ​ - T sin α

0,42 \text{ m}

0, 42 m

0,65 \text{ m}

0, 65 m

N_y=1,2 \text{ kg} \cdot 9,81 \text{ m/s}^2+5,2 \text{ kg} \cdot 9,81 \text{m/s}^2 - 104,84 \text{ N} \cdot \sin 32,869^\circ

N ​ y ​ ​ = 1, 2 k g \cdot 9, 81 m / s ​ 2 ​ ​ + 5, 2 k g \cdot 9, 81 m / s ​ 2 ​ ​ - 104, 84 N \cdot sin 32, 86 9 ​ \circ ​ ​

Lasketaan seinän tukivoiman x- ja y-suuntaisten komponenttien suuruudet.

N_y=G_t +G_k - T \sin \color{CornflowerBlue}\alpha

N ​ y ​ ​ = G ​ t ​ ​ + G ​ k ​ ​ - T sin α

N_y \approx 5,89\text{ N}

N ​ y ​ ​ \approx 5, 89 N

N_x =T \cos \color{CornflowerBlue}\alpha

N ​ x ​ ​ = T cos α

N_x =104,84 \text{ N} \cdot \sin 32,869^{\circ}

N ​ x ​ ​ = 104, 84 N \cdot sin 32, 86 9 ​ \circ ​ ​

N_x \approx 88,1 \text{ N}

N ​ x ​ ​ \approx 88, 1 N

0,42 \text{ m}

0, 42 m

0,65 \text{ m}

0, 65 m

Lasketaan Pythagoraan lauseen avulla seinän tukivoiman suuruus.

N=\sqrt{N_x^2+N_y^2}

N = \sqrt ​ N ​ x ​ 2 ​ ​ + N ​ y ​ 2 ​ ​ ​ ​ ​

N=\sqrt{(88,1 \text{ N})^2+(5,89 \text{ N})^2}

N = \sqrt ​ (88, 1 N) ​ 2 ​ ​ + (5, 89 N) ​ 2 ​ ​ ​ ​ ​

N \approx 88 \text{ N}

N \approx 88 N

Lasketaan tangentin avulla seinän tukivoiman suunta.

\tan \color{CornflowerBlue}\beta = \dfrac{N_y}{N_x}

tan β = \frac{​ N ​ y ​ ​ ​ ​}{​ N ​ x ​ ​ ​}

\tan \color{CornflowerBlue}\beta = \dfrac{5,89 \text{ N}}{88,1 \text{ N}}

tan β = \frac{​ 5 , 8 9 N ​ ​}{​ 8 8 , 1 N ​}

\color{CornflowerBlue}\beta \approx 3,8^\circ

β \approx 3, 8 ​ \circ ​ ​

Newtonin 3. lain mukaan seinän tankoon kohdistama tukivoima on yhtä suuri kuin tangon seinään kohdistama voima, mutta vastakkaisuuntainen.

Vastaus:

Vaijerin jännitysvoiman suuruus on noin 100 N. Seinään kohdistuvan voiman suuruus on noin 88 N ja suunta poikkeaa seinän normaalista noin 3,8 astetta.

Fysiikan ylioppilaskoe S2006

Ratkaisu

Kirjataan lähtöarvot

m_1=1 \text{ kg}, \ x_1=0,32 \text{ m}

m ​ 1 ​ ​ = 1 k g, x ​ 1 ​ ​ = 0, 32 m

m_2=2 \text{ kg}, \ x_2=?

m ​ 2 ​ ​ = 2 k g, x ​ 2 ​ ​ = ?

Piirretään voimakuviot tilanteista.

x_1-x_2 = 0,09 \text{ m}

x ​ 1 ​ ​ - x ​ 2 ​ ​ = 0, 09 m

1 kg:n ja 2 kg:n merkkien välimatka on 9,0 cm.

x_2 = x_1-0,09 \text{ m}

x ​ 2 ​ ​ = x ​ 1 ​ ​ - 0, 09 m

x_2 =0,32 \text{ m}-0,09 \text{ m}

x ​ 2 ​ ​ = 0, 32 m - 0, 09 m

x_2 =0,23 \text{ m}

x ​ 2 ​ ​ = 0, 23 m

Kalavaaka on tasapainossa pyörimisen suhteen molemmissa tilanteissa.

\sum M_A = 0

\sum M ​ A ​ ​ = 0

\color{Salmon}{G_1} x_1 - \color{Salmon}{G_p}(r-x_1)=0

G ​ 1 ​ ​ x ​ 1 ​ ​ - G ​ p ​ ​ (r - x ​ 1 ​ ​) = 0

\sum M_B = 0

\sum M ​ B ​ ​ = 0

\color{Salmon}{G_2} x_2 - \color{Salmon}{G_p}(r-x_2)=0

G ​ 2 ​ ​ x ​ 2 ​ ​ - G ​ p ​ ​ (r - x ​ 2 ​ ​) = 0

\color{Salmon}{G_1} x_1 =\color{Salmon}{G_p}(r-x_1)

G ​ 1 ​ ​ x ​ 1 ​ ​ = G ​ p ​ ​ (r - x ​ 1 ​ ​)

\color{Salmon}{G_p}=\dfrac{ x_1 }{r-x_1}\color{Salmon}{G_1}

G ​ p ​ ​ = \frac{​ x ​ 1 ​ ​ ​ ​}{​ r - x ​ 1 ​ ​ ​} G ​ 1 ​ ​

\color{Salmon}{G_2} x_2 =\color{Salmon}{G_p}(r-x_2)

G ​ 2 ​ ​ x ​ 2 ​ ​ = G ​ p ​ ​ (r - x ​ 2 ​ ​)

\color{Salmon}{G_p}=\dfrac{ x_2 }{r-x_2}\color{Salmon}{G_2}

G ​ p ​ ​ = \frac{​ x ​ 2 ​ ​ ​ ​}{​ r - x ​ 2 ​ ​ ​} G ​ 2 ​ ​

Kalavaa'an paino saatiin esitettyä kahdella tavalla, joten muodostetaan näistä yhtälö.

\color{Salmon}{G_p}=\dfrac{ x_1 }{r-x_1}\color{Salmon}{G_1}

G ​ p ​ ​ = \frac{​ x ​ 1 ​ ​ ​ ​}{​ r - x ​ 1 ​ ​ ​} G ​ 1 ​ ​

\color{Salmon}{G_p}=\dfrac{ x_2 }{r-x_2}\color{Salmon}{G_2}

G ​ p ​ ​ = \frac{​ x ​ 2 ​ ​ ​ ​}{​ r - x ​ 2 ​ ​ ​} G ​ 2 ​ ​

\dfrac{ x_1 }{r-x_1}\color{Salmon}{G_1}=\dfrac{ x_2 }{r-x_2}\color{Salmon}{G_2}

\frac{​ x ​ 1 ​ ​ ​ ​}{​ r - x ​ 1 ​ ​ ​} G ​ 1 ​ ​ = \frac{​ x ​ 2 ​ ​ ​ ​}{​ r - x ​ 2 ​ ​ ​} G ​ 2 ​ ​

\dfrac{ x_1 }{r-x_1}m_1g=\dfrac{ x_2 }{r-x_2}m_2g

\frac{​ x ​ 1 ​ ​ ​ ​}{​ r - x ​ 1 ​ ​ ​} m ​ 1 ​ ​ g = \frac{​ x ​ 2 ​ ​ ​ ​}{​ r - x ​ 2 ​ ​ ​} m ​ 2 ​ ​ g

\dfrac{ x_1 }{r-x_1}m_1=\dfrac{ x_2 }{r-x_2}m_2

\frac{​ x ​ 1 ​ ​ ​ ​}{​ r - x ​ 1 ​ ​ ​} m ​ 1 ​ ​ = \frac{​ x ​ 2 ​ ​ ​ ​}{​ r - x ​ 2 ​ ​ ​} m ​ 2 ​ ​

m_1x_1(r-x_2)=m_2x_2(r-x_1)

m ​ 1 ​ ​ x ​ 1 ​ ​ (r - x ​ 2 ​ ​) = m ​ 2 ​ ​ x ​ 2 ​ ​ (r - x ​ 1 ​ ​)

G=mg

G = m g

: g

\cdot(r-x_1)(r-x_2)

\cdot (r - x ​ 1 ​ ​) (r - x ​ 2 ​ ​)

m_1x_1(r-x_2)=m_2x_2(r-x_1)

m ​ 1 ​ ​ x ​ 1 ​ ​ (r - x ​ 2 ​ ​) = m ​ 2 ​ ​ x ​ 2 ​ ​ (r - x ​ 1 ​ ​)

m_1x_1r-m_1x_1x_2=m_2x_2r-m_2x_1x_2

m ​ 1 ​ ​ x ​ 1 ​ ​ r - m ​ 1 ​ ​ x ​ 1 ​ ​ x ​ 2 ​ ​ = m ​ 2 ​ ​ x ​ 2 ​ ​ r - m ​ 2 ​ ​ x ​ 1 ​ ​ x ​ 2 ​ ​

m_1x_1r-m_2x_2r=m_1x_1x_2-m_2x_1x_2

m ​ 1 ​ ​ x ​ 1 ​ ​ r - m ​ 2 ​ ​ x ​ 2 ​ ​ r = m ​ 1 ​ ​ x ​ 1 ​ ​ x ​ 2 ​ ​ - m ​ 2 ​ ​ x ​ 1 ​ ​ x ​ 2 ​ ​

r(m_1x_1-m_2x_2)=m_1x_1x_2-m_2x_1x_2

r (m ​ 1 ​ ​ x ​ 1 ​ ​ - m ​ 2 ​ ​ x ​ 2 ​ ​) = m ​ 1 ​ ​ x ​ 1 ​ ​ x ​ 2 ​ ​ - m ​ 2 ​ ​ x ​ 1 ​ ​ x ​ 2 ​ ​

r=\dfrac{m_1x_1x_2-m_2x_1x_2}{m_1x_1-m_2x_2}

r = \frac{​ m ​ 1 ​ ​ x ​ 1 ​ ​ x ​ 2 ​ ​ - m ​ 2 ​ ​ x ​ 1 ​ ​ x ​ 2 ​ ​ ​ ​}{​ m ​ 1 ​ ​ x ​ 1 ​ ​ - m ​ 2 ​ ​ x ​ 2 ​ ​ ​}

:(m_1x_1-m_2x_2)

: (m ​ 1 ​ ​ x ​ 1 ​ ​ - m ​ 2 ​ ​ x ​ 2 ​ ​)

Sijoitetaan lukuarvot

r=\dfrac{0,32 \text{ m} \cdot 0,23 \text{ m}(1 \text{ kg} -2 \text{ kg})}{1 \text{ kg} \cdot 0,32 \text{ m}-2 \text{ kg} \cdot 0,23 \text{ m}}

r = \frac{​ 0 , 3 2 m \cdot 0 , 2 3 m ( 1 k g - 2 k g ) ​ ​}{​ 1 k g \cdot 0 , 3 2 m - 2 k g \cdot 0 , 2 3 m ​}

r=\dfrac{x_1x_2(m_1-m_2)}{m_1x_1-m_2x_2}

r = \frac{​ x ​ 1 ​ ​ x ​ 2 ​ ​ ( m ​ 1 ​ ​ - m ​ 2 ​ ​ ) ​ ​}{​ m ​ 1 ​ ​ x ​ 1 ​ ​ - m ​ 2 ​ ​ x ​ 2 ​ ​ ​}

\approx 0,53 \text{ m}

\approx 0, 53 m