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• Notes

$\begin{array}{l}{\text { A large wrecking ball }} \\ {\text { is held in place by two light }} \\ {\text { steel cables (Fig. ES. } 6) \text { . If the }} \\ {\text { mass } m \text { of the wrecking ball is }} \\ {4090 \text { kg, what are (a) the ten- }} \\ {\text { sion } T_{B} \text { in the cable that makes }} \\ {\text { an angle of } 40^{\circ} \text { with the verti- }} \\ {\text { cal and (b) the tension } T_{A} \text { in the }} \\ {\text { horizontal cable? }}\end{array}$

$\sum F_{y}=m a_ y$

$\sum F_ y=0$

$T_{B} \cos 40^{\circ}-m g=0$

$\rightarrow T_{B}=\frac{m g}{\cos 40}=\frac{4090 * 9.8}{\cos 40}$

$∴ T_{B}=5.23 * 10^{4} N$

$\sum F_{x}=m a_{x}$

$T_{B} \sin 40-T_ A=0$

$\rightarrow\left(5.23 * 10^{4}\right) \sin 40=T_ A$

$∴ T_ A=3.36 * 10^{4} N$

$\begin{array}{c}{\text { Find the tension in }} \\ {\text { each cord in Fig. E5.7 if the }} \\ {\text { weight of the suspended object }} \\ {\text { is } w \text { . }}\end{array}$

$\sum \vec{F}=m\vec{a}$

$T_{c}=w$

$\sum F_{y}=0$

$T_{A} \sin 30+T_{B} \sin 45^{\circ}-T_{C}=0 \longrightarrow (1)$

$\sum F_{x}=0$

$T_{B} \cos 45^{\circ}-T_A \cos 30^{\circ}=0 \quad \rightarrow (2)$

* -1

$T_A\text { cos } 30-T_{B} \text { cos } 45^{\circ}=0 \leftarrow (3)$

$T_{A}(\sin 30+\cos 30)-T_{C}=0 \rightarrow T_{A}(\sin 30+\cos 30)=T_{C}$

$∴ T_ A=\frac{T_C}{\sin 30+\cos 30}=0.732 \ T _\mathrm{C}=0.732 \mathrm{w}$

$T_{B}=\frac{T_{A} \cos 30}{\cos 45}=0.897 \mathrm{w}$

$\sum F_{y=0}$

$T_{B} \sin 45-T_{A} \cos 60-T_C=0 \rightarrow (1)$

$\sum F_{x}=0$

$T_{B} \cos 45-T_{A} \sin 60=0 \rightarrow (2)$

(2) * -1

$T_A\text { sin 60 - } T_{B} \text { cos 45 }=0 \longrightarrow \mathbb{(3)}$

$T_{A}(\sin 60-\cos 60)-T_C=0$

$T_{A}=\frac{T_{C}}{\sin 60-\cos 60}=2.73 T_{C}={2.73 w}$

$\rightarrow T_B=\frac{T_A\text { sin 60 }}{\cos 45}=\frac{2.73 \mathrm{w} * \sin 60}{\cos 4 \mathrm{5}}={3.35 \mathrm{w}}$

$\begin{array}{l}{\text { A } 1130-\mathrm{kg} \text { car is held in place by a light cable on a very }} \\ {\text { smooth (frictionless) ramp, as shown in Fig. E5.8. The cable }}\end{array}$

$\sum F_{x}=0$

$T\cos 31-mg \sin 25=0$

$T \cos 31-(1130)(9.8) \sin 25=0$

$T=\frac{(1130 * 9.8) \sin 25}{\cos 31}=5460 \mathrm{N}$

$\sum F_ y=0 \longrightarrow {n}+T \sin 31-m g \cos 25=0$

$\rightarrow n=m g \cos 2 5-T \sin 31 \rightarrow n=(1130)(9.8) \cos 25-5460 \sin 31$

$∴ \quad n=7220 N$