Find the acceleration the cart must have in order for the cereal box at the front of the cart not to fall.

Question

assume that the coefficient of static friction between the cart and the box is 0.38.

The acceleration points to the right.

**after making the free body diagram, figured there was 3 forces acting on the cart, W=mg, the normal force and static friction. after determining the forces in the x and y directions, i calculated the acceleration to be 3.72 m/s^2, but it turns out to be wrong. the right answer is suppose to be 26m/s^2. i dont know how they got this answer.

Answer 1

Consider the box as our system.

Let a = acceleration of the cart

Fix the frame of reference on the cart.

Forces on the box are: -

1. mg vetically downward, where m = mass of the box

2. Normal force N towards right by the cart. (As the cart moves towards right, it pushes the box. That is why N is towards right)

3. Force of friction f by the cart vertically upward. Under gravity, box will tend to fall down. So, the friction is upward.

4. Pseudo force = ma towards left.

Frame of reference is fixed on the cart. In this frame, if we do not want the box to fall, then the acceleration of the box = 0

Taking horizontal component of forces

N = ma-----------------(1)

Taking vertical component

f = mg

Force of friction <= coeff. of static friction * normal force

Therefore, f <= 0.38 N

Or, mg <= 0.38 N---------------(2)

From (1) and (2),

mg <= 0.38ma

Dividing by m,

g <= 0.38a

Or, a>=g/0.38

Therefore, minimum required value of a

= g/0.38 = 9.8/0.38

= 26 m/s^2

Answer 2

Maximum fictional force = μmg

=> maximum acceleration

= μmg / m

= μg

= (0.38) * (9.8)

= 3.724 m/s^2

[If you can post the figure, it can be examined waht is incorrect.]