Grade 11General Physics

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.

larisha sharma

13 Years agoGrade 11

5 Answers

Vikas TU

Approved Tutor Answer13 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

13 Years ago

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

Abhishekh kumar sharma

13 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

12 Years ago

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

F_H²+ F_V² = F².

here F_V= 18N, F = 30N.

so F_H= 24N.

Divy Patel

4 Years ago

37 degree APPROX

, 24 NEWTON..................,.....................................................................................................................................................

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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.

larisha sharma

5 Answers

Vikas TU

Akash Kumar Dutta

Abhishekh kumar sharma

SOURAV MISHRA

Divy Patel

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