Question

A crumb of bread, of mass 0.056 kg, is pulled upon by ants from rival anthills. They exert the following forces: 0.06 N to the north, 0.08 N to the east, 0.02 N to the west, and 0.06 N to the southeast (45° south of east). What is the acceleration of the system? What additional force should be applied by a fifth ant to keep the crumb in equilibrium?

Answer 1

Answer:

a) 1.855m/s^2, 9.71\° to the east-north

b) 0.103N, 9.46° to the west-south

Explanation:

To find the acceleration of the system you can assume that  the forces are applied in a xy plane, where force toward north are directed in the +y direction, and forces to the east in the +x direction. BY taking into account the components of the acceleration for each axis you obtain the following systems of equations:

$0.06N-0.06Nsin(45\°)=ma_y\\\\0.08N-0.02N+0.06cos(45\°)=ma_x$

m: mass of the crumb of bread = 0.056kg

you simplify the equations an replace the values of the mass in order to obtain the acceleration components:

$a_y=\frac{0.017N}{0.056kg}=0.313\frac{m}{s^2}\\\\a_x=\frac{0.102N}{0.056kg}=1.829\frac{m}{s^2}$

$\theta=tan^{-1}(\frac{0.313}{1.829})=9.71\°\\\\a=\sqrt{a_x^2+a_y^2}=\sqrt{(1.829)^2+(0.313)^2}\frac{m}{s^2}=1.855\frac{m}{s^2}$

then, the acceleration of the system has a magnitude of 1.855m/s^2 and a direction of 9.71\° to the east-north

The fifth force must cancel both x an y components of the previous net force, that is:

$0.06N-0.06Nsin(45\°)+F_y=0\\\\F_y=-0.017N\\\\0.08N-0.02N+0.06cos(45\°)+F_x=0\\\\F_x=-0.102N\\\\\phi=tan^{-1}(\frac{-0.017}{-0.102N})=9.46\°$

$F=\sqrt{(0.102)^2+(0.017)^2}N=0.103N$

the, the force needed to reach the equilibrium has a magnitude of 0.103N and a direction of 9.46° to the west-south