Question

The uniform crate has a mass of 50 kg and rests on the cart having an inclined surface. Determine the smallest acceleration that will cause the crate either to tip or slip relative to the cart. What is the magnitude of this acceleration

Answer 1

The answer is below

Step-by-step explanation:

Let g = acceleration due to gravity = 9.81 m/s², x = half of the width of the crate, half of the height of the crate = 0.5 m, a = acceleration of crate, N = force raising the crate

The sum of moment is given as:

`50asin(15)x+50acos(15)0.5=-50(9.81)sin(15)0.5+50(9.81)cos(15)x\ \ \ (1)`

Sum of vertical forces is zero, hence:

`N-50(9.81)cos(15)+50acos(15)=0\ \ \ (2)`

Sum of horizontal force is zero, hence:

`50(9.81)sin(15)-\mu N+50acos(15)=0\\\\50(9.81)sin(15)-0.5 N+50acos(15)=0\ \ \ (3)`

Solving equation 1, 2 and 3 simultaneously gives :

N = 447.8 N, a = 2.01 m/s², x = 0.25 m

x is supposed to be 0.3 m (0.6/2)

The crate would slip because x <0.3 m