Answer:
A) a_max = 4.9 m/s²
B) the metal cabinet will not slip.
C) a_max = 9.8 m/s². The cabinet will slip.
Step-by-step explanation:
A) We are given;
Coefficient of static friction; μ_s = 1
Coefficient of kinetic friction; μ_k = 0.7
Formula for maximum static friction;
F_s = μ_s•N
We are dealing with half of the weight of a flatbed truck. Thus;
N = mg/2
Thus;
F_s(max) = ½μ_s•mg
Now, the maximum acceleration it can achieve on dry concrete will be when the Maximum static friction is reached.
Because after maximum static friction, the cabinet will slip.
Thus; F_s(max) = ma
Therefore,
ma = ½μ_s•mg
m will cancel out to give;
a = ½μ_s•g
a = ½ × 1 × 9.8
a_max = 4.9 m/s²
B) we are told that the coefficient of static friction is now 0.55
Thus, the friction between the metal and the wood is;
F = μ_s•g = 0.55 × 9.8
F = 5.39 N
The acceleration gotten in the first part is less than μ_s•g = 5.39 N, then the metal cabinet will not slip.
C) Now, if we are considering a 4 wheel drive, then we will not divide the mass by 2 and so; N = mg
Like we did in A above;
ma = μ_s•mg
a = μ_s•g
a = 1 × 9.8
a = 9.8 m/s²
Now, this value of a_max is greater than μ_s•g in answer A above.
Thus, a_max > μ_s•g and thus, the cabinet will slip.