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 130.0N while simultandously being hit from the side wigh a force of 4500.0N,what force will the sensor report

Answer 1

Explanation:

Given that,

The dummy is thrown forward with a force of 130 N, $F_1=130\ N$

Side force acting on the dummy, $F_2=4500\ N$

We need to find the force acting on the sensor report. It can be calculated using Pythagoras theorem as :

$F_{net}=\sqrt{F_1^2+F_2^2}$

$F_{net}=\sqrt{130^2+4500^2}$

$F_{net}=4501.87\ N$

So, the net force acting on the sensor report is 4501.87 N. Hence, this is the required solution.

Answer 2

Here in crash test the two forces are acting on the dummy in two different directions

As we know that force is a vector quantity so we need to use vector addition laws in order to find the resultant force on it.

So here two forces are given in perpendicular direction with each other so as per vector addition law we need to use Pythagoras theorem to find the resultant of two vectors

so we can say

$F_{net} = \sqrt{F_1^2 + F_2^2}$

here given that

$F_1 = 130.0 N$

$F_2 = 4500.0 N$

now we will plug in all data in the above equation

$F_{net} = \sqrt{4500^2 + 130^}$

$F_{net} = 4501.9 N$

so it will have net force 4501.9 N which will be reported by sensor