Two blocks of masses m1 and m2, with m1 m2, are placed in contact with each other on a frictionless, horizontal surface, as in Figure 5.12a. A constant horizontal force F is applied to m1 as shown. (A) Find the magnitude of the acceleration of the system.

Question

Two blocks of masses [latex] m_{1} ext { and } m_{2} ext {, with } m_{1}>m_{2} [/latex], are placed in contact with each other on a frictionless, horizontal surface, as in Figure 5.12a. A constant horizontal force F is applied to m1 as shown.

(A) Find the magnitude of the acceleration of the system.

(B) Determine the magnitude of the contact force between the two blocks.

Accepted Answer

Conceptualize the situation using Figure 5.12a and realizing that both blocks must experience the same acceleration because they are in contact with each other and remain in contact throughout the motion. We categorize this as a Newton’s second law problem because we have a force applied to a system and we are looking for an acceleration. To analyze the problem, we first address the combination of two blocks as a system. Because F is the only external horizontal force acting on the system, we have

[latex] \sum F_{x}( ext { system })=F=\left(m_{1}+m_{2} ight) a_{x} [/latex].

(1) [latex] a_{x}=\frac{F}{m_{1}+m_{2}} [/latex].

To finalize this part, note that this would be the same acceleration as that of a single object of mass equal to the combined masses of the two blocks in Figure 5.12a and subject to the same force.

(B) Determine the magnitude of the contact force between the two blocks.

Conceptualize by noting that the contact force is internal to the system of two blocks. Thus, we cannot find this force by modeling the whole system (the two blocks) as a single particle. We must now treat each of the two blocks individually by categorizing each as a particle subject to a net force. To analyze the situation, we first construct a free-body diagram for each block, as shown in Figures 5.12b and 5.12c, where the contact force is denoted by P. From Figure 5.12c we see that the only horizontal force acting on [latex] m_{2} [/latex] [latex] ext { is the contact force } P _{12} ext { (the force exerted by } m_{1} ext { on } m_{2} ext { ), } [/latex] which is directed to the right. Applying Newton’s second law to [latex] m_{2} [/latex] gives

[latex] ext { (2) } \quad \sum F_{x}=P_{12}=m_{2} a_{x} [/latex].

Substituting the value of the acceleration [latex] a_{x} [/latex] given by (1) into (2) gives

(3) [latex] P_{12}=m_{2} a_{x}=\left(\frac{m_{2}}{m_{1}+m_{2}} ight) F [/latex].

To finalize the problem, we see from this result that the contact force [latex] P_{12} [/latex] is less than the applied force F. This is consistent with the fact that the force required to accelerate block 2 alone must be less than the force required to produce the same acceleration for the two-block system.

To finalize further, it is instructive to check this expression for [latex] P_{12} ext { by considering the forces acting on } m_{1} [/latex], shown in Figure 5.12b. The horizontal forces acting on [latex] m_{1} [/latex] are the applied force F to the right and the contact force [latex] P _{21} [/latex] to the left (the force exerted by [latex] m_{2} ext { on } m_{1} [/latex]). From Newton’s third law, [latex] P _{21} ext { is the reaction to } P _{12} ext {, so } P_{21}=P_{12} [/latex] . Applying Newton’s second law to [latex] m_{1} [/latex] gives

(4) [latex] \sum F_{x}=F-P_{21}=F-P_{12}=m_{1} a_{x} [/latex].

[latex] ext { Substituting into (4) the value of } a_{x} ext { from (1), we obtain } [/latex]

[latex] P_{12}=F-m_{1} a_{x}=F-m_{1}\left(\frac{F}{m_{1}+m_{2}} ight)=\left(\frac{m_{2}}{m_{1}+m_{2}} ight) F [/latex].

This agrees with (3), as it must.

Question 5.7: Two blocks of masses m1 and m2, with m1 > m2, are placed ...

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