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I'm not able to visualize the components of $F$ in x and z direction. As I can tell $F_y=300\cos(30^\circ)$.

enter image description here

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    $\begingroup$ The component in $xz$ plane has magnitude $300\sin(30^\circ)$ since $F$ is inclined at $30^\circ$ with $y$. This component is inclined at $30^\circ$ with $z$ axis towards $-x$. So $F_z=300\sin(30^\circ)\cos(30^\circ)\hat k$ and $F_x=-300\sin(30^\circ)\sin(30^\circ)\hat i$. $\endgroup$ Sep 24, 2020 at 23:51
  • $\begingroup$ @ShubhamJohri is it possible to visualize it? $\endgroup$
    – CroCo
    Sep 25, 2020 at 0:06
  • $\begingroup$ It would require some more labelling to explain clearly, but the force vector is applied in the plane of the bent portion containing point A. So you can break the force into one component along an axis parallel to $y$ as you have done, and the other component along the other axis passing through A. Note that this latter component is parallel to $xz$ plane so you need only the angle it makes with the $z$ or $x$ axes to break it into components along $x$ and $z$. $\endgroup$ Sep 25, 2020 at 0:11

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I don't think the drawing is clear. If you want a real answer you should have a three view drawing, which will show where the angles are measured more clearly. When I first looked, the line $30^\circ$ from the force seemed to be perpendicular to the plate, the direction a drill bit would go through to drill the hole. If that is true the force is purely in the $x$ direction. On a second look, the line $30^\circ$ from the force seems to be parallel to $y$. That would support your claim that $F_y=300 \cos(30^\circ)$. I don't understand what the blue line is trying to tell us. I also can't tell whether $F$ has a component in the $z$ direction or not.

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  • $\begingroup$ "you should have a three view drawing" of course this is infeasible for textbooks so I can't blame the author for such difficulty. $\endgroup$
    – CroCo
    Sep 25, 2020 at 0:04
  • $\begingroup$ I don't think a three view is infeasible for textbooks. The author is responsible for making sure the information to solve the problem is clear. Some words describing which direction some of the construction lines are in would help a lot. For example, saying that the line $30^\circ$ from $F$ is parallel to $y$ and $F$ has no $z$ component would go a long way. $\endgroup$ Sep 25, 2020 at 0:21
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You obiously need to do two projections.

See the axis which starts at point A and is perpendicualr to the y axis. Let's call it $v$.

Then $F_v=300 \cdot \sin (30º)$

Now project that one into the x and z components.

$F_x=F_v\cdot cos(150º) = - 300\ N\sin (30º)\cos (30º)$

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