Question

Two astronauts are out in deep space, wrangling a barrel containing tools and supplies. They both exert forces of exactly the same magnitude F1=F2= 820 on the barrel. No other forces act on the barrel.

2 Fi Case A Case B Case C

a) The astronauts could push the barrel im various directions. In which of the cases from the picture above would the net force acting on the barrel be the smallest?

b) In which case from the picture above would the net force acting on the barrel be the largest?

c) finally, astronaut 1 exerts F1 an an angle of 20 degrees from the y axis while astronaut 2 exerts F2 at an angle of 35 degrees from the x axis. Sketch and label the x & y components of both vectors. (Just a basic diagram is needed. nothing complex)

d) calculate the x and y components of both forces.

e) calculate the magnitude of the net force Fnet=F1+F2 acting on the barrel in this final case

Answer 1

Part a) Net force will be smallest in second case.

$F_{net} = \sqrt{F_{1}^2 + F_{2}^2 + 2F_{1}F_{2}cos(180)}$

$F_{net} = \sqrt{F^2 + F^2 + 2FF(-1)}$   (As F1 = F2 = F)

$F_{net} = \sqrt{F^2 + F^2 - 2FF}$

$F_{net} = \sqrt{2F^2 - 2F^2} = 0$

Part b) Net force is highest in first case:

$F_{net} = \sqrt{F_{1}^2 + F_{2}^2 + 2F_{1}F_{2}cos(0)}$

$F_{net} = \sqrt{F^2 + F^2 + 2FF(+1)}$   (As F1 = F2 = F)

$F_{net} = \sqrt{F^2 + F^2 + 2FF}$

$F_{net} = \sqrt{4F^2} = 2F$  (Both the forces will add up)

$F_{net} = \sqrt{4F^2} = 2\times 820 = 1640N$

Part c)

F1 cos(20) F1 F2*sin(35) F2 20 F1 sin(20) 35 F2*cos(35)

Part d)

X-component of F1 = F1 * sin(20) = 840 * 0.34 = 285.6 N

Y-component of F1 = F1 * cos(20) = 840 * 0.94 = 789.6 N

X-component of F2 = F2 * cos(35) = 840 * 0.82 = 688.8 N

Y-component of F2 = F2 * sin(35) = 840 * 0.57 = 478.8 N

Part e ) the angle between the two forces is 75 deg

So net force in this final case is:

$F_{net} = \sqrt{F_{1}^2 + F_{2}^2 + 2F_{1}F_{2}cos(75)}$

$F_{net} = \sqrt{F^2 + F^2 + 2FF(0.26)}$

$F_{net} = \sqrt{F^2 + F^2 + 2F^2(0.26)}$

$F_{net} = \sqrt{F^2(1 + 1 + 2(0.26)) }$

$F_{net} = \sqrt{840^2(2.52) }$

$F_{net} = 1333.46 N$