Question

below are six crates at rest on level surfaces. the crates have different masses and the frictional coefficients (μs,μk) between the crates and the surfaces differ. the same external force is applied to each crate, but none of the crates move. part a rank the crates on the basis of the frictional force acting on them. rank from largest to smallest. to rank items as equivalent, overlap them. view available hint(s)

Explanation:

Step1: Recall friction - force formula

The static - friction force is given by $F_f=\mu_sN$, and the normal force $N = mg$ (on a horizontal surface). So $F_f=\mu_smg$. Calculate the friction force for each crate.
For the first crate with $m_1 = 500\ kg$ and $\mu_{s1}=0.6$, $F_{f1}=\mu_{s1}m_1g=0.6\times500\times g = 300g$.
For the second crate with $m_2 = 750\ kg$ and $\mu_{s2}=0.4$, $F_{f2}=\mu_{s2}m_2g=0.4\times750\times g=300g$.
For the third crate with $m_3 = 250\ kg$ and $\mu_{s3}=0.2$, $F_{f3}=\mu_{s3}m_3g=0.2\times250\times g = 50g$.
For the fourth crate with $m_4 = 750\ kg$ and $\mu_{s4}=0.6$, $F_{f4}=\mu_{s4}m_4g=0.6\times750\times g = 450g$.
For the fifth crate with $m_5 = 600\ kg$ and $\mu_{s5}=0.8$, $F_{f5}=\mu_{s5}m_5g=0.8\times600\times g=480g$.
For the sixth crate with $m_6 = 1500\ kg$ and $\mu_{s6}=0.3$, $F_{f6}=\mu_{s6}m_6g=0.3\times1500\times g = 450g$.

Step2: Rank the friction forces

Comparing the values of $F_{f1},F_{f2},F_{f3},F_{f4},F_{f5},F_{f6}$:
$F_{f5}(480g)>F_{f4}(450g)=F_{f6}(450g)>F_{f1}(300g)=F_{f2}(300g)>F_{f3}(50g)$.

Answer:

From largest to smallest: crate with $m = 600\ kg,\mu_s=0.8$; crate with $m = 750\ kg,\mu_s=0.6$; crate with $m = 1500\ kg,\mu_s=0.3$; crate with $m = 500\ kg,\mu_s=0.6$; crate with $m = 750\ kg,\mu_s=0.4$; crate with $m = 250\ kg,\mu_s=0.2$