+0

# system of blocks

0
367
6
+1832

physics
xvxvxv  Nov 30, 2014

#3
+26971
+10

Like this:

T is tension in string; N is normal reaction force; F is force of frictional resistance.

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Alan  Dec 2, 2014
#1
+26971
+10

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°)

$${\mathtt{1}}{\mathtt{\,\times\,}}{\mathtt{9.8}}{\mathtt{\,\times\,}}\underset{\,\,\,\,^{{360^\circ}}}{{sin}}{\left({\mathtt{40}}^\circ\right)} = {\mathtt{6.299\: \!318\: \!574\: \!932\: \!6}}$$

or approximately 6.3N

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Alan  Dec 1, 2014
#2
+1832
0

Alan

can you make free body digram for each object  ?

xvxvxv  Dec 2, 2014
#3
+26971
+10

Like this:

T is tension in string; N is normal reaction force; F is force of frictional resistance.

.

Alan  Dec 2, 2014
#4
+1832
0

for the free body digram for B

why Tsin40 directid to the left ?

xvxvxv  Dec 3, 2014
#5
+26971
+5

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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Alan  Dec 3, 2014
#6
+1832
0

thank you

xvxvxv  Dec 4, 2014