91Ó°ÊÓ

Two blocks, of masses \(m_{1}\) and \(m_{2}\), moving in the \(x\) -direction are pushed by a force \(F\) as shown in Fig. 3-19. The coefficient of friction between each block and the table is \(0.40 .(a)\) What must be the value of \(F\) if the blocks are to have an acceleration of \(200 \mathrm{~cm} / \mathrm{s}^{2} ?\) How large a force does \(m_{1}\) then exert on \(m_{2}\) ? Use \(m_{1}=300 \mathrm{~g}\) and \(m_{2}=500 \mathrm{~g}\). Remember to work in SI units. The friction forces on the blocks are \(F_{f 1}=0.40 m_{1} g\) and \(F_{f 2}=0.40 m_{2} g\). Take the two blocks in combination as the object for discussion; the horizontal forces on the object from outside (i.e., the external forces on it) are \(F\), \(F_{f 1}\), and \(F_{f 2}\). Although the two blocks do push on each other, those pushes are internal forces; they are not part of the unbalanced external force on the two-mass object. For that object, $$ \pm \sum F_{x}=m a_{x} \text { becomes } F-F_{f 1}-F_{\mathrm{f} 2}=\left(m_{1}+m_{2}\right) a_{x} $$ (a) Solving for \(F\) and substituting known values $$ F=0.40 g\left(m_{1}+m_{2}\right)+\left(m_{1}+m_{2}\right) a_{x}=3.14 \mathrm{~N}+1.60 \mathrm{~N}=4.7 \mathrm{~N} $$ (b) Now consider block \(m_{2}\) alone. The forces acting on it in the \(x\) -direction are the push of block \(m_{1}\) on it (which we represent by \(F_{b}\) ) and the retarding friction force \(F_{12}=0.40 m_{2} g\). Then, for it, $$ \pm \sum F_{x}=m a_{x} \quad \text { becomes } \quad F_{b}-F_{f 2}=m_{2} a_{x} $$ We know that \(a_{x}=2.0 \mathrm{~m} / \mathrm{s}^{2}\) and so $$ F_{b}=F_{\mathrm{f} 2}+m_{2} a_{x}=1.96 \mathrm{~N}+1.00 \mathrm{~N}=2.96 \mathrm{~N}=3.0 \mathrm{~N} $$

Step by step solution

01

Convert mass units to SI

Given masses are in grams, but we need to convert them to kilograms for standard SI units. We have \( m_1 = 300 \text{ g} = 0.3 \text{ kg} \) and \( m_2 = 500 \text{ g} = 0.5 \text{ kg} \).

02

Calculate the force of friction on each block

The force of friction on each block is given by \( F_{f1} = 0.40 \times m_1 \times g \) and \( F_{f2} = 0.40 \times m_2 \times g \), where \( g = 9.8 \text{ m/s}^2 \). Calculate these as: \( F_{f1} = 0.40 \times 0.3 \times 9.8 = 1.176 \text{ N} \) and \( F_{f2} = 0.40 \times 0.5 \times 9.8 = 1.96 \text{ N} \).

03

Calculate the total force needed for acceleration

The sum of forces in the x-direction for the system is \( F - F_{f1} - F_{f2} = (m_1 + m_2) \times a_x \). Substitute the values: \( F - 1.176 - 1.96 = (0.3 + 0.5) \times 2.0 \). This gives \( F - 3.136 = 1.6 \) and solve for \( F \): \( F = 4.736 \text{ N} \).

04

Calculate the force exerted by block 1 on block 2

Isolate block 2 and apply the equation \( F_b - F_{f2} = m_2 \times a_x \) where \( F_b \) is the force exerted by block 1 on block 2. Substitute the known values: \( F_b - 1.96 = 0.5 \times 2.0 \). Solve for \( F_b \): \( F_b = 1.96 + 1.0 = 2.96 \text{ N} \).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Friction

When two surfaces come into contact, they experience resistance to their motion, known as friction. In this scenario, the blocks are interacting with the table surface, leading to a frictional force opposing their movement. The frictional force can be calculated using the equation:

• \( F_f = \mu \cdot m \cdot g \)

where \( \mu \) represents the coefficient of friction, \( m \) is the mass of the block, and \( g \) is the acceleration due to gravity, typically \( 9.8 \text{ m/s}^2 \).
In our exercise, both blocks encounter a frictional force, with \( \mu = 0.40 \). Thus, the forces are

• \( F_{f1} = 0.40 \times m_1 \times g \)
• \( F_{f2} = 0.40 \times m_2 \times g \)

This is crucial for determining the total force needed to achieve a certain acceleration.

Acceleration

Acceleration refers to the rate of change in the velocity of an object. It is determined by dividing the net force acting on the object by its mass. In mathematical terms, this is expressed as:

• \( a = \frac{F_{net}}{m} \)

For our problem, the desired acceleration is \( 200 \text{ cm/s}^2 \), which is equivalent to \( 2.0 \text{ m/s}^2 \).
We need to calculate the net force that considers individual block friction to achieve this acceleration for our combined block system. This demands that the applied force \( F \) overcomes both the frictional forces \( F_{f1} \) and \( F_{f2} \) to push the blocks forward.
Understanding acceleration in this context helps explain how forces are required to set objects in motion and change their velocity.

Force Calculation

Calculating the force necessary for moving the blocks involves accounting for the friction they experience. The total force is given by:

• \( F = F_{f1} + F_{f2} + (m_1 + m_2) \times a \)
• \( F = 0.40 \times g \times (m_1 + m_2) + (m_1 + m_2) \times a \)

By substituting the appropriate values for gravity \( g \), mass \( m_1 = 0.3 \text{ kg} \), and \( m_2 = 0.5 \text{ kg} \), as well as the acceleration, the force \( F \) is determined to be approximately \( 4.7 \text{ N} \).
Furthermore, analyzing the forces between the blocks as they interact, particularly the force applied by block 1 on block 2, is crucial. This is represented by the force \( F_b \), calculated as

• \( F_b = F_{f2} + m_2 \times a \)

This results in \( 3.0 \text{ N} \) for this scenario.
These calculations highlight how related forces interact and influence the outcome when multiple objects are involved.