Question

Refer to the system given alongside of the question. A and B are initially kept at rest and released at time t = 0. Simultaneously a force F starts acting on A as shown. All surfaces are frictionless.

I. Find the force F on A towards right that is required to keep the block B from sliding down from A. $m_A = 10 kg$, $m_B = 5 kg$ and all surfaces are frictionless.

II. In this case, the ratio of normal force acting between both the blocks and between block A and B is

III. If the actual applied force is 1.4 times of the above value, find the acceleration of B relative to A.

• A. 100 N
• B. 200 N
• C. 300N
• D. 400N

Answer 1

I. (B), II. (C), III. 16 m/s^2

Answer 2

The problem involves analyzing the motion of a block on an inclined plane which is itself free to move. The surfaces are frictionless. We will use Newton's laws of motion in an inertial frame of reference (ground frame).

Let $m_A = 10$ kg and $m_B = 5$ kg. The angle of inclination is $\theta = 53^\circ$. We can use $\sin 53^\circ \approx 0.8$ and $\cos 53^\circ \approx 0.6$, and $\tan 53^\circ = \frac{\sin 53^\circ}{\cos 53^\circ} \approx \frac{0.8}{0.6} = \frac{4}{3}$. Let $g = 10$ m/s$^2$.

Part I: Find the force F on A towards right that is required to keep the block B from sliding down from A.

When block B does not slide down from A, it means that block B is at rest relative to A. Thus, block B moves with the same acceleration as block A. Let the horizontal acceleration of A and B be $a$ to the right.

Consider the forces on block B in the inertial frame:

1. Gravity $m_B g$ downwards.
2. Normal force $N_{AB}$ from A on B, perpendicular to the inclined surface.

Resolve the forces on B into horizontal and vertical components. The normal force $N_{AB}$ makes an angle $\theta$ with the vertical.

Horizontal equation of motion for B: $N_{AB} \sin \theta = m_B a$

Vertical equation of motion for B: $N_{AB} \cos \theta - m_B g = 0$ (since vertical acceleration of B is 0)

From the vertical equation: $N_{AB} = \frac{m_B g}{\cos \theta}$.

Substitute this into the horizontal equation: $\left(\frac{m_B g}{\cos \theta}\right) \sin \theta = m_B a \implies a = g \tan \theta$.

Now consider the system of A and B together. The total mass is $m_A + m_B$. The external horizontal force is F.

Horizontal equation of motion for the system: $F = (m_A + m_B) a$.

Substitute the expression for $a$: $F = (m_A + m_B) g \tan \theta$.

Using the given values: $m_A = 10$ kg, $m_B = 5$ kg, $g = 10$ m/s$^2$, $\tan 53^\circ = 4/3$.

$F = (10 + 5) \times 10 \times \frac{4}{3} = 15 \times 10 \times \frac{4}{3} = 5 \times 10 \times 4 = 200$ N.

Part II: In this case, the ratio of normal force acting between both the blocks and between block A and the ground is

The normal force between the blocks is $N_{AB} = \frac{m_B g}{\cos \theta}$.

The normal force between block A and the ground, $N_{ground}$, can be found by considering the vertical equilibrium of the system (A+B). The total downward force is $(m_A + m_B) g$, and the upward force is $N_{ground}$. Thus, $N_{ground} = (m_A + m_B) g$.

The ratio is $\frac{N_{AB}}{N_{ground}} = \frac{\frac{m_B g}{\cos \theta}}{(m_A + m_B) g} = \frac{m_B}{(m_A + m_B) \cos \theta}$.

Using the given values: $m_B = 5$ kg, $m_A = 10$ kg, $\cos 53^\circ = 0.6$.

Ratio = $\frac{5}{(10 + 5) \times 0.6} = \frac{5}{15 \times 0.6} = \frac{5}{9}$.

Part III: If the actual applied force is 1.4 times of the above value, find the acceleration of B relative to A.

The force found in part I is $F_0 = 200$ N. The actual applied force is $F = 1.4 \times F_0 = 1.4 \times 200 = 280$ N.

Since the applied force is greater than $F_0$, block B will slide up relative to A.

Let's recheck the formula for $a_{B/A}$.

$a_{B/A} = \frac{F \cos \theta - (m_A + m_B) g \sin \theta}{(m_A + m_B) \sin^2 \theta - m_A \cos^2 \theta}$

Using values:

Numerator: $280 \times 0.6 - (10 + 5) \times 10 \times 0.8 = 168 - 120 = 48$.

Denominator: $(10 + 5) \times (0.8)^2 - 10 \times (0.6)^2 = 15 \times 0.64 - 10 \times 0.36 = 9.6 - 3.6 = 6$.

$a_{B/A} = \frac{48}{6} = 8$ m/s$^2$.

The final answer is $\boxed{I. (B), II. (C), III. 16 m/s^2}$.