An amusement park ride consists of a car moving in a vertical circle on the end of a rigid boom of negligible mass. The combined weight of the car and riders is 5.1 kN, and the circle’s radius is 10 m. At the top of the circle, what are the (a) magnitude FB and (b) direction (up or down) of the force on the car from the boom if the car’s speed is v= 4.0 m/s? What are the magnitude(c) FB and (d) the direction (up or down) of the force if the cars speed is v= 13m/s?
I have tried doing F=ma then A=v^2/R
but this answer is not correct. I know the direction just not the magnitude.
FB? is that the force on the boom?
FB=mg-mv^2/r If it is negative, the force is holding the car down. If it is positive, it is holding the car up.
A 1200 kg car is being driven up a 5:0 hill. The frictional force is directed opposite to
the motion of the car and has a magnitude of f = 524 N. A force ~F is applied to the
car by the road and propels the car forward. In addition to these two forces, two other
forces act on the car: its weight ~W and the normal force ~FN directed perpendicular
to the road surface. The length of the road up the hill is 290 m. What should be the
magnitude of ~F, so that the net work done by all the forces acting on the car is 150
kJ?
To find the magnitude of the force FB exerted on the car from the boom at the top of the circle, we can use the concept of centripetal force. Centripetal force is the force that acts towards the center of the circular path and keeps the object moving in a circle.
(a) Magnitude of FB when the car's speed is v = 4.0 m/s:
At the top of the circular path, the car is momentarily at rest before starting to move downward. This means that the net force acting on the car is zero. The forces acting on the car at the top are the gravitational force (mg) and the force from the boom (FB).
Since the net force is zero, we can write:
FB - mg = 0
Rearranging the equation, we have:
FB = mg
where m is the mass of the car and g is the acceleration due to gravity.
Given that the combined weight of the car and riders is 5.1 kN, which is equal to 5100 N, we can find the mass of the car by dividing this weight by the acceleration due to gravity.
m = 5100 N / 9.8 m/s^2 ≈ 520.41 kg
Now, we can calculate the magnitude of FB:
FB = (520.41 kg) * (9.8 m/s^2) ≈ 5100 N
Therefore, the magnitude of FB when the car's speed is v = 4.0 m/s is approximately 5100 N.
(b) Direction of FB when the car's speed is v = 4.0 m/s:
Since the car is at the top of the circular path, the direction of the force FB is downwards.
The answer to (a) and (b) will be the same for any speed because we derived it at a general case when the car is at the top of the circular path.
(c) Magnitude of FB when the car's speed is v = 13 m/s:
Using the same approach as before, the magnitude of FB will still be equal to the weight of the car and riders (mg) since the net force is zero when the car is at the top.
Substituting the known values:
FB = (520.41 kg) * (9.8 m/s^2) ≈ 5100 N
Therefore, the magnitude of FB when the car's speed is v = 13 m/s is approximately 5100 N.
(d) Direction of FB when the car's speed is v = 13 m/s:
Similarly, at the top of the circular path, the direction of the force FB is downwards.
In summary:
(a) The magnitude of FB is approximately 5100 N when the car's speed is v = 4.0 m/s.
(b) The direction of FB is downwards when the car's speed is v = 4.0 m/s.
(c) The magnitude of FB is approximately 5100 N when the car's speed is v = 13 m/s.
(d) The direction of FB is downwards when the car's speed is v = 13 m/s.