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

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

the figure given was of a cart that has a cereal box hanging off the front of it.

in my Free Body Diagram i had normal force pointing up, mg pointing down, frictional force pointing to the left.

it was given that the acceleration points to the right.

Update:

FIND THE MAXIMUM ACCELERATION THE CART MUST HAVE IN ORDER FOR THE CEREAL BOX AT THE FRONT OF THE CART NOT TO FALL.

Relevance
• Anonymous

Zero is one valid answer to the problem.

MAXIMUM acceleration is a different question.

• Anonymous
5 years ago

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

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

Note: - It is minimum acceleration and not maximum.