A cord is wrapped around the rim of a solid uniform wheel 0.280 m in radius and of mass 7.80 kg. A steady horizontal pull of 34.0 N to the right is exerted on the cord, pulling it off tangentially from the wheel. The wheel is mounted on frictionless bearings on a horizontal axle through its center. Required:
a. Compute the angular acceleration of the wheel
b. Compute the acceleration of the part of the cord that has already been pulled off the wheel.
c. Find the magnitude of the force that the axle exerts on the wheel.
d. Find the direction of the force that the axle exerts on the wheel.
e. Which of the answers in parts A, B, C, and D would change if the pull were upward instead of horizontal?

Solution :

Given :

Mass, m = 7.80 kg

Radius, r = 0.280 m

Force, F = 34 N

a). St = I x a

b). The acceleration of the part of the chord is

    = 31.13 x 0.280

c). Magnitude of the force is =

                                               = 83.73 N

d). The direction of force is = arctan (7.8 x 9.81)/34

                                             = 2.25

e). The answers in part (c) and (d) will change when the pull of force is given in the upward direction instead of giving in the horizontal direction

if the forces acting on an object are unbalanced, the object will accelerate in the direction of the net force on the object.

explanation:

this is best explained by having an understanding of newton's second law, which states that f=ma. you see, when the forces are unbalanced, net force (sum of all forces) is not zero. therefore making it necessary for the object to have an acceleration in the direction of the net force.