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mechanics (1 Viewer)
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scaryshark09
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is the answer C
Taking the unit vectors i and j as upwards and to the right, the forces acting are:
\,\boldsymbol{i} + (F\sin{\alpha} + N\cos{\alpha} + F - mg)\,\boldsymbol{j})
Since the object is stationary, this sum must be the zero vector. It then follows, from the i components, that:
and, from the j components, that:
 + N\cos{\alpha} &= mg \\ F( 1+ \sin{\alpha}) + \cfrac{F\cos{\alpha}}{\sin{\alpha}} \times \cos{\alpha} &= mg \qquad \text{using the above result for $N$} \\ F\left( 1+ \sin{\alpha} + \cfrac{\cos^2{\alpha}}{\sin{\alpha}}\right) &= mg \\ F\left( \sin{\alpha} + \sin^2{\alpha} + \cos^2{\alpha}\right) &= mg\sin{\alpha} \\ F &= \cfrac{mg\sin{\alpha}}{ \sin{\alpha} + 1} \qquad \text{after using $\sin^2{\alpha} + \cos^2{\alpha} = 1$} \\ \text{And, as $\alpha \approx 37^\circ \implies \sin{\alpha} \approx \cfrac{3}{5}$}: \qquad \qquad F &\approx mg \times \cfrac{\cfrac{3}{5}}{\cfrac{3}{5} + 1} \\ &\approx \cfrac{3mg}{8} \end{align*})
and the answer is (D).
Note, substituting the value 37 degrees gives F = 0.375708... mg, which is far closer to (D) than to any of the other options.
- a weight force of W = -mgj
- the tension in the string of T = +Fj
- the friction force of
- the normal force of
Since the object is stationary, this sum must be the zero vector. It then follows, from the i components, that:
and, from the j components, that:
Note, substituting the value 37 degrees gives F = 0.375708... mg, which is far closer to (D) than to any of the other options.
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Also regarding SHM say you have a point B to the right of the origin and the particle is initially moving right. Is the time taken to go from O to B the same time to go from B to O? ( this low key is a dumb question but just checking)
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This is true from looking at the sine / cosine graph. I think it should also be true whether the particle initially moving left or if B is to the left of O.