Answer:
The force does the ceiling exert on the hook is 269.59 N
Explanation:
Applying the second Newton law:
F = m*a
From the attached diagram, the net force in object 1 is:
[tex]m_{1} a=T_{1} -W_{1}[/tex]
In object 2:
[tex]m_{2} a=W_{2} -T_{2}[/tex]
Adding the two equations:
[tex]m_{2} a+m_{1} a=T_{1} -W_{1} +W_{2} -T_{2} \\m_{1} =\frac{W_{1} }{g} \\m_{2} =\frac{W_{2} }{g} \\Replacing\\T_{2}-T_{1}=W_{2} -W_{1} -(\frac{W_{1} }{g} +\frac{W_{2} }{g} )a[/tex] (eq. 1)
The torque:
[tex]\tau =I\alpha[/tex]
Where
I = moment of inertia
α = angular acceleration
If the linear acceleration is
[tex]a=r\alpha \\\alpha =\frac{a}{r} \\I=\frac{1}{2} mr^{2} \\\tau =\frac{mra}{2}[/tex]
Torque due the tension is equal:
[tex]\tau =r(T_{2} -T_{1} )[/tex]
Substituting torque, mass, in equation 1, the expression respect the acceleration is:
[tex]a=\frac{g*(W_{2}-W_{1})}{W_{1}+W_{2} +\frac{W}{2} }[/tex]
Where
W₁ = 75 N
W₂ = 125 N
W = 80 N
[tex]a=\frac{9.8*(125-75)}{75+125+\frac{80}{2} } =2.04m/s^{2}[/tex]
The net force is:
[tex]F_{n} =F-W-T_{1} -T_{2}\\0=F-W-W_{1} (\frac{a}{g} +1)-W_{2} (1-\frac{a}{g})\\F=W+W_{1} +W_{2} +\frac{a}{g} (W_{1} -W_{2} )\\F=80+75+125+\frac{2.04}{9.8} (75-125)\\F=269.59N[/tex]
We have that for the Question "What force does the ceiling exert on the hook "
It can be said that
From the question we are told
Two weights are connected by a very light cord that passes over an 80.0N frictionless pulley of radius 0.300m
Assuming the two weights are 125N and 75N
[tex]\sum F = ma\\\\125-75 = (\frac{125}{g} + \frac{75}{g})a\\\\a = \frac{1}{4}g[/tex]
For the separate system,
[tex]T - 75 = \frac{75}{g}a\\\\T = \frac{75}{g}*\frac{1}{4}g + 75\\\\T = 93.75N[/tex]
Therefore,
[tex]F_{ceiling} - W_{pulley} - 2T = 0\\\\F_{ceiling} = 80 + 2*93.75\\\\= 267.5N[/tex]
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