Question

a 20 kg sig is pulled by a horizontal force such that the single rope holding the sign make an angle of 21 degree with the verticle assuming the sign is motionless find the magnitude of the tension in the ropend the magnitude of the horizontal force?

Answer 1

T= 210.15 N

F= 75.31 N

Step-by-step explanation:

Let the tension in string be T newton.

According to the question

⇒T×cos21°= mg

⇒T= mg/cos21°

⇒T=20×9.81/cos21

⇒T= 210.15 N

now, the magnitude of horizontal force

F= Tsin21°

⇒F= 210.15×sin21°

=75.31 N