A force of magnitude 7.50 {\rm N} pushes three boxes with masses m_1 = 1.30\,{\rm{kg}},m_2 = 3.20\,{\rm{kg}}, and m_3 = 4.90\,{\rm{kg}}, as shown in the figure(Figure 1) . (Ignore friction.)
Find the magnitude of the contact force between boxes 1 and 2, and
Find the magnitude of the contact force between boxes 2 and 3.
no
To find the magnitudes of the contact forces between the boxes, we can use Newton's second law, which states that the force acting on an object is equal to the mass of the object multiplied by its acceleration.
Let's denote the contact force between boxes 1 and 2 as F12, and the contact force between boxes 2 and 3 as F23.
1. Magnitude of the contact force between boxes 1 and 2 (F12):
The force applied to box 1 is 7.50 N, and since there is no friction, this force is equal to the contact force between boxes 1 and 2. Therefore, F12 = 7.50 N.
2. Magnitude of the contact force between boxes 2 and 3 (F23):
To find this force, we need to consider the net force acting on box 2. The force applied to box 2 is 7.50 N, and there is also a contact force from box 1, which is F12. The net force acting on box 2 is the vector sum of these two forces.
Since box 2 is being pushed from both sides, we can represent the net force as:
Net force on box 2 = F12 - F23
According to Newton's second law, this net force is equal to the mass of box 2 multiplied by its acceleration, which is assumed to be the same for all the boxes (a):
F12 - F23 = m2 * a
Substituting the known values, we can solve for F23:
7.50 N - F23 = 3.20 kg * a
We can use the same equation to find the acceleration (a) by considering the net force acting on box 1:
Net force on box 1 = F12 = m1 * a
Substituting the known values:
7.50 N = 1.30 kg * a
Now we have two different expressions for the acceleration (a), so we can equate them and solve for F23:
1.30 kg * a = 3.20 kg * a
Since the accelerations are the same for all the boxes, they cancel out, leaving:
7.50 N = F12 - F23
We already know that F12 = 7.50 N, so we can substitute this value:
7.50 N = (7.50 N) - F23
Solving for F23:
F23 = 0 N
Therefore, the magnitude of the contact force between boxes 2 and 3 is 0 N.
To find the magnitudes of the contact forces between the boxes, we need to first determine the net force acting on each box.
Let's start with finding the net force on box 1.
Net force on box 1 (F_net1) = Force applied (7.50 N) - Contact force between boxes 1 and 2 (F_12)
Next, we calculate the net force on box 2.
Net force on box 2 (F_net2) = Contact force between boxes 1 and 2 (F_12) - Contact force between boxes 2 and 3 (F_23)
Finally, we determine the net force on box 3.
Net force on box 3 (F_net3) = Contact force between boxes 2 and 3 (F_23)
Since there are no other forces acting on the boxes (assuming no friction), the net force is equal to the mass times the acceleration for each box, according to Newton's second law (F = ma).
For box 1:
F_net1 = m_1 * a_1
For box 2:
F_net2 = m_2 * a_2
For box 3:
F_net3 = m_3 * a_3
Since the boxes are pushed together and move as a system, they all have the same acceleration (a_1 = a_2 = a_3 = a).
By equating the net force equations with the mass-acceleration equations, we can solve for the magnitudes of the contact forces.
For box 1:
m_1 * a = Force applied - F_12 --> F_12 = Force applied - m_1 * a
For box 2:
m_2 * a = F_12 - F_23 --> F_23 = F_12 - m_2 * a
For box 3:
m_3 * a = F_23 --> F_23 = m_3 * a
Since both expressions for F_23 are equal to each other, we can substitute the second expression into the first one to find F_12:
F_12 = Force applied - m_1 * a - m_2 * a = Force applied - (m_1 + m_2) * a
Now we can substitute the given values to find the magnitudes of the contact forces.
m_1 = 1.30 kg
m_2 = 3.20 kg
m_3 = 4.90 kg
Force applied = 7.50 N
To find the magnitude of the contact force between boxes 1 and 2 (F_12):
F_12 = Force applied - (m_1 + m_2) * a
Substituting the values:
F_12 = 7.50 N - (1.30 kg + 3.20 kg) * a
To find the magnitude of the contact force between boxes 2 and 3 (F_23):
F_23 = F_12 - m_2 * a
Substituting the values:
F_23 = F_12 - 3.20 kg * a