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Sagot :
According to the problem, we know that
• The angle of the roof is 35°.
,• The coefficient of static friction is 0.51.
,• Mazie's mass is 42 kg.
It's important to know that the force involved in this problem is Mazie's weight.
[tex]F_{\text{weight}}=mg[/tex]Given that the roof is inclined by 35°, we get the rectangular components of the weight.
[tex]\begin{gathered} F_{Wy}=mg\cos 35 \\ F_{Wx}=mg\sin 35 \end{gathered}[/tex]Usually, the vertical component uses sine and the horizontal component uses cosine, however, due to the inclination of the roof, these trigonometric reasons change.
Now, using Newton's Second Law, we express the sum of forces about each component.
[tex]\begin{gathered} \Sigma F_x=ma \\ F_{Wx}+F_f=ma \end{gathered}[/tex]This means the forces that affect Mazie horizontally are the static friction and the horizontal component of her weight. Let's use the friction force definition to include in the expression
[tex]mg\sin \theta+\mu mg\cos \theta=ma[/tex]Note that the normal force used in the expression of friction is actually equivalent to the vertical component of the weight. Now, the problem is asking for the maximum angle so she doesn't slip, which means we have to find an angle where there's no acceleration in the system because that would be mean she is not going to fall. So, a = 0 and let's replace the other variables to solve for theta
[tex]\begin{gathered} 42\operatorname{kg}\cdot9.81m/s^2\cdot\sin \theta+0.51\cdot42\operatorname{kg}\cdot9.81m/s^2\cdot\cos \theta=m\cdot0 \\ 412.02\cdot\sin \theta+210.13\cdot\cos \theta=0 \\ 412.02\cdot\sin \theta=-210.13\cdot\cos \theta \\ 412.02\cdot\frac{\sin\theta}{\cos\theta}=-210.13 \\ 412.02\tan \theta=-210.13 \\ \tan \theta=-\frac{210.13}{412.02} \\ \theta\approx27 \end{gathered}[/tex]This means that the maximum angle the rood could have without her slipping is 27°, approximately.
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