Question

61. Three rugby players are pulling horizontally on es attached to a box, which remains stationary. Player 1 exerts a force F1 equal to 100 N at an angle 01 equal to 60° with respect to the +x direction (Figure 4-39). Player 2 exerts a force F2 equal to 200 N at an angle 02 equal to 37° with respect to the +x direction. The view in the figure is from above. Ignore friction and note that gravity plays no role in this problem! SSM Figure 4-39 Problem 61 00N 2 37 (a) Find the force F3 exerted by Player 3. State your an- swer by giving the components of F3 in the directions perpendicular to and parallel to the positive x direction. (b) Redraw the diagram and add the force F3 as care- fully as you can. (c) Player 3's rope breaks, and Player 2 adjusts by pulling with F2 equal to 150 N at the same angle as before. In which direction is the acceleration of the box, relative to the direction shown? (d) In part (c),

Answer 1

a)

let the force applied by the third player is

$\vec F_3= F_x\hat x+F_y \hat y$

along the x-axis, taking the components of F1 and F2

$\\F_{X-net}= |\vec F_1|cos\theta_1+ |\vec F_2|cos \theta _2 +F_{3x}\\ \\ F_{X-net} =200N cos(37^o)+100Ncos (60^0) =209.72 N+ F_{3x}$

but since the ball is in equilibrium, the net x- force must be 0

$\\ \\ F_{X-net} =200N cos(37^o)+100Ncos (60^0) =209.72 N+ F_{3x} =0 \\ \\ \implies F_{3x}= -209.72 N$

similarly for the y-axis

$\\F_{Y-net}= |\vec F_2|sin\theta_2- |\vec F_1|sin \theta _1 +F_{3y}\\ \\ F_{Y-net} =200N sin(37^o)-100Nsin (60^0) =33.76N+ F_{3y}$

applying the equilibrium condition on the y-axis,

$\\ \\ F_{Y-net} =33.76N+ F_{3y} \\ 0=33.76N+ F_{3y} \\ F_{3y} =-33.76N$

hence

$\vec F_3=-209.72\hat x-33.76N \hat y$

___________________________________________________

b)

the angle made by this force vector with negative x-axis

$\\tan\theta_3= \frac{|F_{3y}|}{|F_{3x}|}=\frac{33.76N}{209.72N}\\ \\ \theta _3=tan^{-1}(\frac{33.76N}{209.72N})= 9.14^o$

F3 9. 14 2 looM

__________________________________________

c)

Now , after breaking the rope, the force applied by player 3 is zero. , the magnitude of F2 is now 150 N.

the box is no longer in equilibrium. the accelertion will be in the direction of the applied force.

So we need to find the direction of the net force now.

$\\F_{Y-net}= |\vec F_2|sin\theta_2- |\vec F_1|sin \theta _1 \\ \\ F_{Y-net} =150N sin(37^o)-100Nsin (60^0) =3.7NN$

on the x-axis

$\\F_{X-net}= |\vec F_1|cos\theta_1+ |\vec F_2|cos \theta _2 \\ \\ F_{X-net} =150N cos(37^o)+100Ncos (60^0) =169.8 N$

we need to find the angle made by the positive x-axis

$\\tan\theta_3= \frac{|F_{Y-net}|}{|F_{X0net}|}=\frac{3.7N}{169.8N}\\ \\ \theta _3=tan^{-1}(\frac{3.7N}{169.8N})= 1.25^o$