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a force of 30N is inclined at an angle of theta to the horizontal.If its vertical component is 18N. find theta and horizontal component.

a force of 30N is inclined at an angle of theta to the horizontal.If its vertical component is 18N. find theta and horizontal component.

Grade:11

5 Answers

Vikas TU
14149 Points
10 years ago

30sinthetha = 18

sinthetha = 18/30

              = 3/5

 

thetha = 37''

horizontal component = 30costhetha = 30cos37

                                                    = 30 x 4/5

                                                    = 24 N

 

plz approve!

 

Akash Kumar Dutta
98 Points
10 years ago

vert comp=18=30sin x
sinx=3/5
x=37 deg

Abhishekh kumar sharma
34 Points
10 years ago

30sinthetha = 18

sinthetha = 18/30

              = 3/5

 

thetha = 37''''

horizontal component = 30costhetha = 30cos37

                                                    = 30 x 4/5

                                                    = 24 N

 

plz approve!

 

SOURAV MISHRA
37 Points
10 years ago

the answer can be obtained by the application of pythagoras theorem as well.

FH2 + FV2  = F2.

here F= 18N, F = 30N.

so F= 24N.

Divy Patel
13 Points
one year ago
37 degree APPROX
 , 24 NEWTON..................,.....................................................................................................................................................

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