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A sphere of mass 3.0104kgis suspended from a cord. A steady horizontal breeze pushes the sphere so that the cord makes a constant angle role="math" localid="1657172689218" 37oofwith the vertical. (a) Find the push magnitude and (b) Find the tension in the cord.

Short Answer

Expert verified

(a) The magnitude of pushing force2.2103N.

(b) The tension in the cord3.710-3N.

Step by step solution

01

Given information

  • The mass of the sphere is,m=3.010-4kg.
  • The angle made by the cord with the vertical is,=37o.
02

Understanding the concept of force

Force acting on the object is equal to the mass multiplied by the acceleration of the object. It is called Newton鈥檚 second law of motion. The free-body diagram shows different forces acting on the object along with their directions.

Using the free-body diagram, Newton鈥檚 second law and vector resolution concept, the pushing force on the sphere and tension in the cord can be found.

Formulae:

Fnet=ma

Here,Fnetis the net force acting on the object, mis the mass of the object, andais the acceleration of the object.

03

Draw the free body diagram

04

a) Calculate the magnitude of pushing force

By applying Newton鈥檚 second law along x axis as,

Fnet=ma

Due to pushing force the sphere moves in upward direction as shown in the figure. Consider the sign convention according to the direction of forces acting on the sphere as shown in the figure.

By applying Newton鈥檚 second law along x axis as

localid="1657176304348" Fnet=maF.cos-mg.sin=0F.cos=mg.sinF=3.010-4kg9.8m/s2sin(37o)cos(37o)=2.210-3N

Hence, the magnitude of pushing force is 2.210-3N.

05

b) Calculate the tension in the chord T

The tension in the cord is acting in the upward direction, hence applying Newton鈥檚 second law in the vertical direction. The vertical acceleration is zero.

T-mg.cos-F.sin=0T=mg.cos+F.sin=(3.010-4)kg9.8m/s2cos(37o)+(2.210-3Nsin(37o))=3.710-3N

Hence, the tension in the chord is 3.710-3N.

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