Answer:
The answer is below
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:
[tex]50asin(15)x+50acos(15)0.5=-50(9.81)sin(15)0.5+50(9.81)cos(15)x\ \ \ (1)[/tex]
Sum of vertical forces is zero, hence:
[tex]N-50(9.81)cos(15)+50acos(15)=0\ \ \ (2)[/tex]
Sum of horizontal force is zero, hence:
[tex]50(9.81)sin(15)-\mu N+50acos(15)=0\\\\50(9.81)sin(15)-0.5 N+50acos(15)=0\ \ \ (3)[/tex]
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
