+33615 The frictional force must exactly balance the horizontal force supplied by the string. The tension in the string angled at 40° to the vertical is the same as the weight of C (because the pulley is frictionless). The horizontal component of this is just 1kg*9.8m/s2*sin(40°)
$${\mathtt{1}}{\mathtt{\,\times\,}}{\mathtt{9.8}}{\mathtt{\,\times\,}}\underset{\,\,\,\,^{\textcolor[rgb]{0.66,0.66,0.66}{360^\circ}}}{{sin}}{\left({\mathtt{40}}^\circ\right)} = {\mathtt{6.299\: \!318\: \!574\: \!932\: \!6}}$$
or approximately 6.3N
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+33615 Like this:
T is tension in string; N is normal reaction force; F is force of frictional resistance.
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+33615 On B, mass A is applying a force that is trying to pull B to the left. This is balanced by the pull from the sloping string over the pulley, which has a horizontal component of Tsin(40°). Because the system is stationary, the pull to the left must have the same magnitude as the pull to the right. (Note: the vertical component of the pull from the sloping string exactly balances the weight of B).
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