Question 5.13: Acceleration of Two Connected Objects When Friction Is Prese...

Conceptualize Imagine what happens as \overrightarrow{F} is applied to the block. Assuming \overrightarrow{F} is large enough to break the block free from static friction but not large enough to lift the block, the block slides to the right and the ball rises.

Categorize We can identify forces and we want an acceleration, so we categorize this problem as one involving two particles under a net force, the ball and the block. Because we assume that the block does not rise into the air due to the applied force, we model the block as a particle in equilibrium in the vertical direction.

Analyze First draw force diagrams for the two objects as shown in Figures 5.21b and 5.21c. Notice that the string exerts a force of magnitude T on both objects. The applied force \overrightarrow{F} has x and y components F cos θ and F sin θ, respectively. Because the two objects are connected, we can equate the magnitudes of the x component of the acceleration of the block and the y component of the acceleration of the ball and call them both a. Let us assume the motion of the block is to the right.

Apply the particle under a net force model to the block in the horizontal direction:

(1)   \sum F_x=F \cos \theta-f_k-T=m_2 a_x = m_2 a

Because the block moves only horizontally, apply the particle in equilibrium model to the block in the vertical direction:

(2)   \sum F_y=n+F \sin \theta-m_2 g=0

Apply the particle under a net force model to the ball in the vertical direction:

(3)   \sum F_y=T-m_1 g=m_1 a_y=m_1 a

Solve Equation (2) for n:

Substitute n into f_k=\mu_k n from Equation 5.10:

(4)   f_k=\mu_k\left(m_2 g-F \sin \theta\right)

Substitute Equation (4) and the value of T from Equation (3) into Equation (1):

Solve for a:

(5)   a=\frac{F\left(\cos \theta+\mu_k \sin \theta\right)-\left(m_1+\mu_k m_2\right) g}{m_1+m_2}

Finalize The acceleration of the block can be either to the right or to the left depending on the sign of the numerator in Equation (5). If the velocity is to the left, we must reverse the sign of f_k in Equation (1) because the force of kinetic friction must oppose the motion of the block relative to the surface. In this case, the value of a is the same as in Equation (5), with the two plus signs in the numerator changed to minus signs.

What does Equation (5) reduce to if the force \overrightarrow{F} is removed and the surface becomes frictionless? Call this expression Equation (6). Does this algebraic expression match your intuition about the physical situation in this case? Now go back to Example 5.10 and let angle u go to zero in Equation (5) of that example. How does the resulting equation compare with your Equation (6) here in Example 5.13? Should the algebraic expressions compare in this way based on the physical situations?