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I am given a problem where I am to pick up the prism shown below by touching only the upper two faces. The coefficient of friction with each face is 0.4 and the mass of the prism is 30grams. What is the minimum force I must apply to each face in order to lift the prism?

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The one thing I am stuck on are the directions of the force I apply with my fingers. I am not sure whether

a)the forces exerted by my fingers act horizontally inwards (which is slightly confusing me; theoretically, should horizontal forces not be able to counteract the downwards weight of the prism? And yet my calculations in this case give a force required of 0.61N)

b)or whether they act exactly normal to the surface (in which case i get 0.22N)

c)or whether they can act directly up the slope straight away, or at least with some upwards inclination to counteract the weight.

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  • $\begingroup$ The horizontal frictional force is related to the normal applied force. Use that concept. Your fingers should be assumed to be applying force perpendicular to the face of the prism. $\endgroup$ – Kalpak Gupta Apr 9 '17 at 2:23
  • $\begingroup$ @KalpakGupta could you please explain why I should assume my fingers apply force perpendicular to the faces? $\endgroup$ – Meep Apr 9 '17 at 10:21
  • $\begingroup$ 1) Hold an object. Keep your fingers horizontal. If you were applying force not perpendicular to it, it would be translating. But that object is stationary! This means the force must be perpendicular to the surface. 2) Imagine no gravity and no friction. Can you hold the prism without translation? The answer is no, because in that case the forces (perpendicular to the surface) do not cancel out because the prism faces are not parallel. $\endgroup$ – Kalpak Gupta Apr 9 '17 at 12:21
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Your fingers apply forces which have components $F, N$ which are parallel and perpendicular to the faces of the prism. These components are related through the usual relation $F=\mu N$.

In order to lift the prism the resultant of all the applied forces must equal the weight of the prism. So you will have to find the vertical components of $F$ and $N$. The horizontal components cancel out.

enter image description here

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  • $\begingroup$ Hi Sammy Gerbil, thanks for your reply. The problem is that then when I resolve for the net force to cancel the weight of the prism, I have an expression in both F and the angle made by the force with the prism faces (rather than using components F and N, I assumed the forces F were applied at an angle $\theta $ above the normals to the prism faces) so to find F I would need to know the angle of application of the forces. I differentiated my expression for the sum of vertical components with respect to the angle, ad thankfully I obtained a stationary point that was a maximum (so that the $\endgroup$ – Meep Apr 9 '17 at 10:13
  • $\begingroup$ force needed would be a minimum), however this angle is 82.5degrees in the case of this problem!) which is almost parallel up the slope of he face! Does this seem reasonable? I don't seem to have very of intuition for these things... I also doubt whether this approach is correct a) because this seems awefully long for a question designed to take only 5minutes in an exam and b) Kalpak Guitars comment suggests to apply the force perpendicular to the faces of the prism. Perhaps this is indeed the full solution if I were able to apply the force at any angle I wished, however maybe in reality $\endgroup$ – Meep Apr 9 '17 at 10:18
  • $\begingroup$ when I pick up an object I only apply forces perpendicular to the surfaces I touch? I cannot rationalise why this would be the case though... $\endgroup$ – Meep Apr 9 '17 at 10:19
  • $\begingroup$ You must apply normal and friction forces. If there is no friction force then you cannot pick up the prism, unless it is inverted, because N has no upward component in this orientation. You must also apply friction force F : the upward vertical component of F balances the downward component of N and the weight W. ... Apply forces $F,N$ parallel and perpendicular to each face, where $F=\mu N$. Write another equation to balance vertical forces. Solve for $F, N$. Resultant applied force $R=\sqrt{F^2+N^2}$ at angle $\alpha$ to prism face where $\tan\alpha=N/F$. $\endgroup$ – sammy gerbil Apr 9 '17 at 16:37
  • $\begingroup$ There is no need to differentiate. Minimum means that the prism does not accelerate, the total force on the prism is zero. ... The question is ambiguous : it does not make it clear if it is asking for the value of N or R. Kalpak is thinking only of N, but F is a necessary consequence of N, and without F you cannot lift the prism. $\endgroup$ – sammy gerbil Apr 9 '17 at 16:46

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