f7-1:if p = 200n,determine the friction that developed between the 50kg crate and the ground.the coefficient…

Explanation:

Step1: Calculate the normal force

The weight of the crate $W = mg$, where $m = 50kg$ and $g=9.81m/s^{2}$. So $W=50\times9.81 = 490.5N$. The vertical - component of the force $P$ is $P_y=P\times\frac{3}{5}$, with $P = 200N$, so $P_y=200\times\frac{3}{5}=120N$. The normal force $N$ on the crate is $N = W - P_y=490.5- 120=370.5N$.

Step2: Calculate the maximum static friction force

The formula for the maximum static friction force is $F_{s,\max}=\mu_sN$. Given $\mu_s = 0.3$ and $N = 370.5N$, then $F_{s,\max}=0.3\times370.5 = 111.15N$. The horizontal - component of the force $P$ is $P_x=P\times\frac{4}{5}$, with $P = 200N$, so $P_x=200\times\frac{4}{5}=160N$. Since $P_x=160N>F_{s,\max}=111.15N$, the crate is in motion and the friction force is the kinetic - friction force. But if we assume we are only interested in the static - friction situation before motion occurs, the friction force $F$ that develops is equal to the maximum static friction force when the crate is on the verge of moving. So $F = 111.15N$.

Answer:

$111.15N$