the sensor in the torso of a crash-test dummy records the magnitude and direction of the net force acting on the dummy. if the dummy is thrown forward with a force of 130.0 N while simultaneously being hit from side with a force of 4500.0 N, what force will the sensor report?

Question

what force will the seat belt have to exert on the dummy to hold the dummy in the seat?

Answer 1

sqrt (130^2 + 4500^2)

tan angle from straight ahead = 4500/130

seat belt equal and opposite

Answer 2

4500/130= 34.6N

Answer 3

sqrt (130^2 + 4500^2) = 4501.8 N

tan(angle) (4500.0/130.0)

Answer 4

you subtract 130 from 4500, and you get 4470 :)

Answer 5

sqrt (130^2 + 4500^2) = 4501.9N
tan(angle) = (opp/adj)
Angle = (tan^-1)(4500/130)
Angle = 88.3 degrees to the side of forward

Answer: -4501.9N @-88.3 degrees to the side of forward

(answer due to opposite of what sensor reads)

Answer 6

5. The sensor in the torso of a crash-test dummy records the magnitude and direction of the net force acting on the dummy. If the dummy is thrown forward with a force of 130.0 N while simultaneously being hit from the side with a force of 4500.0 N, what force will the sensor report?​

Answer 7

Fx=130 N
Fy=4500 N
Fnet=fx+fy
4500+130=4.630
Therefor force of sensor report = 4.630