Question

Three boxes in contact rest side-by-side on a smooth, horizontal floor. Their masses are 5.0-kg, 3.0-kg, and 2.0-kg, with the 3.0-kg box in the center. A force of 50 N pushes on the 5.0- kg box, which pushes against the other two boxes. (a) Draw the free-body diagrams for each of the boxes. (b) Write Newton’s equation for each mass along the horizontal direction. (c) What magnitude force does the 3.0-kg box exert on the 5.0- kg box? (d) What magnitude force does the 3.0-kg box exert on the 2.0kg box?

Answer 1

Answer:

(a)Look at the attached graphic

(b)

(b)-1 Equation 1  : m1= 5kg

       50-F1= 5 *a

(b)-2 Equation 2 : m2= 3kg

        F1-F2= 3 *a

(b)-3 Equation 3 : m3= 2kg

         F2 = 2*a  

(c) F1 =25 N

(d) F2 =10 N

Explanation:

We apply Newton's second law:

∑F = m*a (Formula 1)

∑F : algebraic sum of the forces in Newton (N)

m : mass in kilograms (kg)

a : acceleration in meters over second square (m/s²)

(a) Draw the free-body diagrams for each of the boxes

Look at the attached graphic

(b) Write Newton’s equation for each mass along the horizontal direction.

Data: m1=  5.0-kg ,m2= 3.0-kg , ,m3= 2.0-kg

Look m1 free-body diagram:

∑Fx = m1*a

50-F1= 5 *a Equation 1

Look m2 free-body diagram:

∑Fx = m2*a

F1-F2= 3 *a Equation 2

Look m3 free-body diagram:

∑Fx = m3*a

F2 = 2*a     Equation 3

(c) What magnitude force does the 3.0-kg box exert on the 5.0- kg box?

Look Free body diagram of the mass set

∑Fx = m*a   m= m1+m2+m3= 5+3+2 = 10 kg

50 = 10*a

a= 50/10 = 5 m/s²

We replace a = 5 m/s² in the equation 1:

50-F1= 5 *5

50-25= F1

F1 = 25 N

(d) What magnitude force does the 3.0-kg box exert on the 2.0kg box?

We replace a= 5 m/s² in the equation 3

F2 = 2*5 = 10 N