Q 1 :    

 A 1 kg mass is suspended from the ceiling by a rope of length 4 m. A horizontal force 'F' is applied at the midpoint of the rope so that the rope makes an angle of 45° with respect to the vertical axis as shown in the figure. The magnitude of F is:         [2024]

  • 102N

     

  • 1 N

     

  • 110×2N

     

  • 10 N

     

(4)

As the system is in equilibrium, therefore

T1cos45°=mg

T1sin45°=F

 tan45°=FmgF=mg=10 N

 



Q 2 :    

A body of mass 1 kg is suspended with the help of two strings making angles as shown in figure. Magnitude of tensions T1 and T2, respectively, are (in N):          [2025]

  • 5, 53

     

  • 53, 5

     

  • 53, 53

     

  • 5, 5

     

(2)

T1sin60°+T2sin30°=mg           ... (i)

T1cos60°=T2cos30°                       ... (ii)

T1=3T2                                                 ... (iii)

From (i) and (iii)

3T2×32+T22=mg

T2=mg2=5 N

and T1=3mg2=53 N



Q 3 :    

A body of mass m is suspended by two strings making angles θ1 and θ2 with the horizontal ceiling with tensions T1 and T2 simultaneously. T1 and T2 are related by T1=3T2. The angle θ1 and θ2 are          [2025]

  • θ1=30°, θ2=60° with T2=3mg4

     

  • θ1=60°, θ2=30° with T2=mg2

     

  • θ1=45°, θ2=45° with T2=3mg4

     

  • θ1=30°, θ2=60° with T2=4mg5

     

(2)

T1sinθ1+T2sinθ2=mg

T1cosθ1=T2cosθ2

3T2cosθ1=T2cosθ2

3cosθ1=cosθ2

For θ1=60°, θ2=30°

3T2·32+T2·12=mg T2=mg2