Respuesta :
Answer:
a) The force is 1056.93 N
b) If the sprocket radius is 2.88 cm, the force is 1662.47 N
c) The combined mass is 102.08 kg
Explanation:
a) The net torque is:
[tex]\tau =I\alpha =mr^{2} \alpha[/tex]
Where
m = 1.82 kg
r = 32 cm = 0.32 m
α = 4.5 rad/s²
Replacing:
[tex]\tau =1.82*0.32^{2} *4.5=0.839Nm[/tex]
The force is:
[tex]\tau =rF-(147*0.32)\\F=\frac{\tau +(147*0.32)}{r}[/tex]
Where
r = 4.53 cm = 0.0453 m
Replacing:
[tex]F=\frac{0.839+47.04}{0.0453} =1056.93N[/tex]
b) If r = 2.88 cm = 0.0288 m, replacing in the previous equation:
[tex]F=\frac{0.839+47.04}{0.0288} =1662.47N[/tex]
c) The combined mass is:
[tex]m=\frac{F}{a} =\frac{147}{0.32*4.5} =102.08kg[/tex]
a) The force is 1056.93 N.
b) If the sprocket radius is 2.88 cm, the force is 1662.47 N.
c) The combined mass is 102.08 kg.
Calculation of the force, radius, mass:
a. Since the net torgue is
T = mr^a
Here,
m = 1.82 kg
r = 32 cm = 0.32 m
α = 4.5 rad/s²
So, T = 1.82*0.32^2*4.5
= 0.839 Nm
Now the force is
T = rF - (147*0.32)
F = T+(147*0.32)/r
= 0.839+47.04/0.0453
= 1056.93 N
b. Since r = 2.88 cm = 0.0288 m
F = 0.839+47.04/0.0288
= 1662.47 N
c. Here the combined mass should be
m = F/s
= 1.47/0.32*4.5
= 102.08 kg
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