NLM BYU 22

Step 1: Define Forces on the Wedge

Let the acceleration of the wedge be $a$ towards the right. The string exerts a force $F$ on the pulley horizontally to the right, and a tension force $F$ downwards along the incline. The block $m$ exerts a normal force $N$ on the wedge perpendicular to the incline.

Horizontal Equation of Motion for Wedge ($M$):

The net horizontal force is the sum of:

• Force from string (horizontal part): $F$ (Right)
• Force from string (inclined part acting on pulley): Horizontal component is $-F \cos \theta$ (Left)
• Normal force from block $m$: Horizontal component is $N \sin \theta$ (Right)

$$F – F \cos \theta + N \sin \theta = M a \quad \dots (1)$$

Step 2: Analyze Block m (in Wedge’s Frame)

In the non-inertial frame of the wedge (accelerating right with $a$), a pseudo force $ma$ acts on block $m$ to the left.

Forces perpendicular to the incline:

• Normal force $N$ (outward)
• Component of Gravity: $mg \cos \theta$ (inward)
• Component of Pseudo force: Since pseudo force is horizontal (left), its component perpendicular to the slope pushes the block OUT OF the incline: $ma \sin \theta$.

Since there is no motion perpendicular to the incline:

$$N = mg \cos \theta – ma \sin \theta \quad \dots (2)$$

Step 3: Solve for Acceleration

Substitute equation (2) into equation (1):

$$F(1 – \cos \theta) + (mg \cos \theta – ma \sin \theta) \sin \theta = M a$$ $$F(1 – \cos \theta) + mg \sin \theta \cos \theta – ma \sin^2 \theta = M a$$

Rearrange to group terms with $a$:

$$F(1 – \cos \theta) + mg \sin \theta \cos \theta = M a + ma \sin^2 \theta$$ $$a (M + m \sin^2 \theta) = F(1 – \cos \theta) + mg \sin \theta \cos \theta$$