A sled that has a mass of 8 kg is pulled at a 50 degree angle with a force of 20 N. The force of friction acting on the sled is
2.4 N. The free-body diagram shows the forces acting on the sled.
What is the acceleration of the sled and the normal force acting on it, to the nearest tenth?
What is the acceleration of the sled and the normal force acting on it, to the nearest tenth?
1 answer:
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Answer:
Explanation:
The question is ;
3.80 m-long, 500. kg steel beam extends horizontally from the point
where it has been bolted to the framework of a new building under
construction. A 67.0 kg construction worker stands at the far end of
the beam. What is the magnitude of the torque about the point where the beam is bolted into place?
Solution:
Here two torques are in action
a) torque T1 due to weight of worker at the edge of the beam - at center of the beam
b) torque T2 due to weight of the uniform beam - at the point where the beam is bolted
So we first calculate the torque produced due to weight of the worker;
We can see that;
Distance of worker from the center of the beam = 1.9m
Mass of the worker = 67 kg
Value of g= 9.8 m/sec2
T1=force × distance from point of rotation
Here force is weight of the worker which is = mass × g=67×9.8=656.6 N
So the torque is
T1= 656.6×1.9=12225.892 Nm
or
T1 = 12225.892 Nm
Now torque by the beam itself ;
Length of the beam from its center point to the bolt point = 1.9 m
Weight force of beam acting at center point= mass× g= 500×9.8=4900 N
Torque T2 at bolt point by the beam weight = weight of beam × length of beam from its center to the bolt point = 4900×1.9=9310 Nm
T2=9310Nm
So total torque = T1+T2= 12225.892+9310=21565.892 Nm
