Question

Two weights are connected by a very light cord that passes over an 80.0Nfrictionless pulley of radius 0.300m. The pulley is a solid uniform disk and is supported by a hook connected to the ceiling. What force does the ceiling exert on the hook

Answer 1

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:

$m_{1} a=T_{1} -W_{1}$

In object 2:

$m_{2} a=W_{2} -T_{2}$

Adding the two equations:

$m_{2} a+m_{1} a=T_{1} -W_{1} +W_{2} -T_{2} \\m_{1} =\frac{W_{1} }{g} \\m_{2} =\frac{W_{2} }{g} \\Replacing\\T_{2}-T_{1}=W_{2} -W_{1} -(\frac{W_{1} }{g} +\frac{W_{2} }{g} )a$  (eq. 1)

The torque:

$\tau =I\alpha$

Where

I = moment of inertia

α = angular acceleration

If the linear acceleration is

$a=r\alpha \\\alpha =\frac{a}{r} \\I=\frac{1}{2} mr^{2} \\\tau =\frac{mra}{2}$

Torque due the tension is equal:

$\tau =r(T_{2} -T_{1} )$

Substituting torque, mass, in equation 1, the expression respect the acceleration is:

$a=\frac{g*(W_{2}-W_{1})}{W_{1}+W_{2} +\frac{W}{2} }$

Where

W₁ = 75 N

W₂ = 125 N

W = 80 N

$a=\frac{9.8*(125-75)}{75+125+\frac{80}{2} } =2.04m/s^{2}$

The net force is:

$F_{n} =F-W-T_{1} -T_{2}\\0=F-W-W_{1} (\frac{a}{g} +1)-W_{2} (1-\frac{a}{g})\\F=W+W_{1} +W_{2} +\frac{a}{g} (W_{1} -W_{2} )\\F=80+75+125+\frac{2.04}{9.8} (75-125)\\F=269.59N$