Unit 2: Force and Translational Dynamics
Weightage : 20-15%
A person is cleaning a wall by pushing a brush against the wall, as shown in the figure. The person pushes the brush upward while it moves upward with an increasing speed.
Which of the following free-body diagrams best represents the forces acting on the brush?
Correct Answer: \( \boxed{\mathrm{D}} \)
The brush experiences four forces:
• Its weight acts downward.
• The wall exerts a normal force horizontally away from the wall.
• The person’s hand pushes the brush into the wall, producing an equal and opposite horizontal force toward the wall.
• Since the brush moves upward relative to the wall, the person must apply an upward force that is greater than the downward weight for the brush to accelerate upward.
Because the brush is moving upward with an increasing speed, its acceleration is upward. Therefore, the net vertical force must be upward.
The horizontal forces balance each other, so the leftward and rightward force vectors are equal in magnitude. The upward force is greater than the downward weight, resulting in an upward net force.
Diagram D correctly represents these forces: the diagonal applied force has a rightward component that balances the wall’s normal force and an upward component that exceeds the downward weight.
Therefore, the correct answer is (D).
A block of mass \(m\) is initially moving on a rough horizontal surface as shown in Figure 1. Figure 2 shows the four forces acting on the block.
Which of the following best describes the motion of the block at the instant shown?
Correct Answer: \( \boxed{\mathrm{C}} \)
The velocity vector in Figure 1 points to the right, so the block is moving to the right.
From the free-body diagram, the normal force \(N\) balances the weight \(mg\), so there is no vertical acceleration.
Horizontally, the frictional force \(f\) acting to the left is larger than the applied force \(F\) acting to the right. Therefore, the net horizontal force is directed to the left.
By Newton’s Second Law,
\(\sum F_x=ma\),
so the acceleration is to the left, opposite the direction of motion.
When an object’s acceleration is opposite its velocity, its speed decreases.
Therefore, the block is moving to the right and slowing down.
Therefore, the correct answer is (C).
Blocks \(A\) and \(B\) of masses \(m_A\) and \(m_B\), respectively, are attached to an ideal pulley to form the Atwood machine shown above. The system is released from rest, and \(m_B>m_A\).
Which of the following free-body diagrams could correctly represent the forces acting on the two blocks?
Correct Answer: \( \boxed{\mathrm{D}} \)
Since \(m_B>m_A\), block \(B\) accelerates downward while block \(A\) accelerates upward.
For block \(A\),
\(T-m_Ag=m_Aa\),
so the tension must be greater than the weight, \(T>m_Ag\).
For block \(B\),
\(m_Bg-T=m_Ba\),
so the weight must be greater than the tension, \(m_Bg>T\).
Because the pulley and string are ideal, the tension has the same magnitude on both sides of the pulley.
Therefore, the correct free-body diagram must show:
• Block \(A\): upward tension larger than downward weight.
• Block \(B\): downward weight larger than upward tension.
• Equal tension forces on both blocks.
Only Diagram D satisfies all of these conditions.
Therefore, the correct answer is (D).
