To solve for the acceleration of the crate, let's break the problem into key components. We can apply Newton's Second Law of Motion in both the horizontal and vertical directions and account for the forces involved.

Given data:

• Mass of the crate, $m = 20 \, \text{kg}$
• Tension in the handle, $T = 40 \, \text{N}$
• Coefficient of friction, $\mu = 0.14$
• Angle of the handle, $\tan \alpha = \frac{3}{4}$, so $\alpha = \tan^{-1}\left(\frac{3}{4}\right)$
• Gravitational acceleration, $g = 9.8 \, \text{m/s}^2$

Step 1: Resolve forces in the horizontal and vertical directions

1. Vertical direction (up and down):

   • The vertical forces acting on the crate are the weight of the crate, the normal force, and the vertical component of the tension force.
   • The weight of the crate is $W = mg = 20 \times 9.8 = 196 \, \text{N}$ downward.
   • The vertical component of the tension in the rope is $T_v = T \sin \alpha$.
   • The normal force $N$ acts upward.

   Using the trigonometric relation $\sin \alpha = \frac{3}{5}$ (since $\tan \alpha = \frac{3}{4}$, and using the Pythagorean identity $3^2 + 4^2 = 5^2$): $T_v = 40 \times \frac{3}{5} = 24 \, \text{N}$

   The normal force is balanced by the weight of the crate and the vertical component of the tension: $N = W - T_v = 196 - 24 = 172 \, \text{N}$

2. Horizontal direction (left and right):

   • The horizontal forces acting on the crate are the horizontal component of the tension force and the frictional force opposing the motion.
   • The frictional force is given by $f_{\text{friction}} = \mu N = 0.14 \times 172 = 24.08 \, \text{N}$.
   • The horizontal component of the tension is $T_h = T \cos \alpha$.
   • Using $\cos \alpha = \frac{4}{5}$, we get: $T_h = 40 \times \frac{4}{5} = 32 \, \text{N}$

Step 2: Apply Newton’s Second Law

Now, we apply Newton's Second Law in the horizontal direction to find the acceleration.

The net force $F_{\text{net}}$ in the horizontal direction is the difference between the horizontal component of the tension and the frictional force: $F_{\text{net}} = T_h - f_{\text{friction}} = 32 - 24.08 = 7.92 \, \text{N}$

Using Newton's Second Law $F_{\text{net}} = ma$, we can solve for the acceleration $a$: $a = \frac{F_{\text{net}}}{m} = \frac{7.92}{20} = 0.396 \, \text{m/s}^2$

Final Answer:

The acceleration of the crate is approximately $0.40 \, \text{m/s}^2$.

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Would you like further details on the steps, or have any questions?

Related Questions:

1. How does the frictional force change if the coefficient of friction were higher?
2. What would happen to the acceleration if the mass of the crate were doubled?
3. How would the result change if the tension in the rope was increased?
4. What role does the angle of inclination play in determining the normal force and friction?
5. How can we calculate the work done by the frictional force during the crate’s motion?

Tip: When solving problems involving friction, remember to break the forces into components and use Newton's Second Law in each direction.