Question

A dockworker loading a 20kg crate onto a ship finds that if he applies a pushing force of 70n at an angle of 40 degrees below the horizontal , the crate just begins to move. what is the coefficient of static friction between the crate and the dock?

Can you please help with a FBD and explain if there is acceleration or not?

Answer 1

Answer 2

Determining the Coefficient of Static Friction (μₛ)

Given:

• Mass of crate ($m$) = 20 kg

• Applied force ($F_{\text{push}}$) = 70 N at 40° below the horizontal

• The crate just begins to move (implying $a=0$ at the threshold of motion).

• 
  

---

Step 1: Free Body Diagram (FBD) and Force Resolution

1. Forces Acting on the Crate:

• Applied Force ($F_{\text{push}}$): Resolve into horizontal ($F_x$) and vertical ($F_y$) components.

  $$F_x=F_{\text{push}}\cos (40\mathrm{°})=70\cos (40\mathrm{°})\approx 53.62\text{N}$$$$F_y=F_{\text{push}}\sin (40\mathrm{°})=70\sin (40\mathrm{°})\approx 44.99\text{N}(\text{directed downward})$$

• Weight ($W$):

  $$W=mg=20\times 9.81=196.2\text{N}(\text{directed downward})$$

• Normal Force ($N$): Balances the total downward forces.

  $$N=W+F_y=196.2+44.99=241.19\text{N}$$

• Static Friction ($f_s$): Opposes impending motion. At the threshold, $f_s={\mu }_{s}N$.

2. Key Observation:
   Since the crate just begins to move, the horizontal forces are balanced (no acceleration):

   $$F_x=f_s⟹53.62={\mu }_s\times 241.19$$

---

Step 2: Solve for μₛ

$${\mu }_s=\frac{F_x}{N}=\frac{53.62}{241.19}\approx 0.222$$

---

 Answer:

The coefficient of static friction between the crate and the dock is ${\mu }_s=0.22$