Question

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?

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