Solving Forces: Balancing Force with -43.6°

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Homework Statement



RZiwx.png


The Attempt at a Solution


define RHS as +ve
in a part, i have calculared that force by conveyor is +1.962N

in b part,
Aer7X.png

then, to balance force, 4.91sin@+1.962=1.962cos@ => @=-43.6'
i an confused about the negative sign... what is my problem?
 
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Your free body diagram is missing a normal force due to the belt
 
JaWiB said:
Your free body diagram is missing a normal force due to the belt

that's true, thx
 
help =[
 

Thread 'Please tell me where I am going wrong in this integral'

##|\Psi|^2=\frac{1}{\sqrt{\pi b^2}}\exp(\frac{-(x-x_0)^2}{b^2}).## ##\braket{x}=\frac{1}{\sqrt{\pi b^2}}\int_{-\infty}^{\infty}dx\,x\exp(-\frac{(x-x_0)^2}{b^2}).## ##y=x-x_0 \quad x=y+x_0 \quad dy=dx.## The boundaries remain infinite, I believe. ##\frac{1}{\sqrt{\pi b^2}}\int_{-\infty}^{\infty}dy(y+x_0)\exp(\frac{-y^2}{b^2}).## ##\frac{2}{\sqrt{\pi b^2}}\int_0^{\infty}dy\,y\exp(\frac{-y^2}{b^2})+\frac{2x_0}{\sqrt{\pi b^2}}\int_0^{\infty}dy\,\exp(-\frac{y^2}{b^2}).## I then resolved the two...
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Thread 'Number of quantum accessible states of a particle given T, N?'

It's given a gas of particles all identical which has T fixed and spin S. Let's ##g(\epsilon)## the density of orbital states and ##g(\epsilon) = g_0## for ##\forall \epsilon \in [\epsilon_0, \epsilon_1]##, zero otherwise. How to compute the number of accessible quantum states of one particle? This is my attempt, and I suspect that is not good. Let S=0 and then bosons in a system. Simply, if we have the density of orbitals we have to integrate ##g(\epsilon)## and we have...
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