Help on a few questions regarding Newton's laws

Question

I'm really struggling with these problems. If anyone could help me out with these it would be greatly appreciated:

 

1.) 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 5.2 kN, and the radius of the circle is 12 m. At the top of the circle, (a) what is the force FB on the car from the boom (using the minus sign for downward direction) if the car's speed is v = 5.7 m/s? (b) What is FB if v = 17 m/s?

 

2.) An airplane is flying in a horizontal circle at a speed of 450 km/h (see the figure). If its wings are tilted at angle θ = 39° to the horizontal, what is the radius of the circle in which the plane is flying? Assume that the required force is provided entirely by an “aerodynamic lift” that is perpendicular to the wing surface. Use g = 9.8 m/s2.

 

3.) A student of weight 657 N rides a steadily rotating Ferris wheel (the student sits upright). At the highest point, the magnitude of the normal force Upper F(Normal) on the student from the seat is 599 N. (a) What is the magnitude of Upper F(Normal) at the lowest point? If the wheel's speed is doubled, what is the magnitude FN at the (b) highest and (c) lowest point?

 

4.) A puck of mass m = 1.50 kg slides in a circle of radius r = 15.0 cm on a frictionless table while attached to a hanging cylinder of mass M = 2.80 kg by a cord through a hole in the table. What speed keeps the cylinder at rest?

 

5.) In the figure here, a box of Cheerios (mass mC = 1.1 kg) and a box of Wheaties (mass mW = 2.8 kg) are accelerated across a horizontal surface by a horizontal force applied to the Cheerios box. The magnitude of the frictional force on the Cheerios box is 2.9 N, and the magnitude of the frictional force on the Wheaties box is 4.3 N. If the magnitude of F is 12.6 N, what is the magnitude of the force on the Wheaties box from the Cheerios box?

 

Thanks!

Answer 1

1) The amusement park ride

Weight of the car and riders = 5.2 kN,

Radius of the circle = 12 m.

Speed of car = (a) 5.7 m/s

(b) 17 m/s

Required F_B ?

============================

On the top of the circle, the two acting forces are:

Gravity----------------------> downward

Centripetal escape ------> upwards

The gravity force is given as weight as 5.2 kN. This would determine the mass m = weight (kN)/ g (m/s²), where g is the gravity acceleration.

Therefore, the mass of the moving parts = (5.2 x 1000 N)/(9.8 m/s²)

Or

m = 530.61 kilograms -------------------------(1)

The centripetal force is a function of the radius of rotation, square of speed, and mass of parts and is expressed as follows:

Centripetal force = m v² / R

Where, v is the speed of parts, (a) 5.7 m/s and (b) 17 m/s

m is the mass = 530.61 kilograms,

and

R is the radius of the circle = 12 m

That gives two values of centripetal forces depending on two values of speeds.

(a) F_c ( at v = 5.7 m/s) = m v² / R

= 530.61 x (5.7²) /12

= 1436.62 N ----------------------(2)

(b ) F_c ( at v = 17 m/s) = m v² / R

= 530.61 x (5.7²) /12

= 12778.86 N---------------------(3)

The net force on the beam of the circle = Fc - weight

(a) At speed 5.7 m/s:

Net Force = 1436.62 N (upward) - 5200 N (downward weight)

= -3763.38 N (downward)

F_b ≈ -3.8 kN

(b) At speed 17 m/s:

Net Force = 12778.86 N (upward) - 5200 N (downward weight)

= 7578.86 N (upward)

F_b ≈ 7.6 kN