Question

if a 50 n block is resting on a steel table with a coefficient of static friction $mu_{s}=0.74$, then what minimum force is required to move the block?
____ n
do not include units in your answer.
enter the answer

Explanation:

Step1: Identify the formula for static - friction force

The formula for the maximum static - friction force is $F_s=\mu_sN$, where $\mu_s$ is the coefficient of static friction and $N$ is the normal force.

Step2: Determine the normal force

The block is resting on a horizontal table, so the normal force $N$ equals the weight of the block. Given that the weight of the block $W = 50$ N, so $N = 50$ N.

Step3: Calculate the maximum static - friction force

Substitute $\mu_s=0.74$ and $N = 50$ N into the formula $F_s=\mu_sN$. We get $F_s=0.74\times50$.
$F_s = 37$

Answer:

37