- #1

- 295

- 2

This is what I did...

4. A 2.00 kg block is held in equilibrium on an incline of angle θ = 70° by a horizontal force vector F applied in the direction shown in the figure below. If the coefficient of static friction between block and incline is μ

_{s}= 0.300, determine the following.

(a) the minimum value of F

Step 1) Draw Free Body Diagram

Step 2) Determine forces in x and y direction

**F**

_{x}f

_{s}

F cos 70

-mg sin 70

**F**

_{y}n

-F sin 70

-mg cos 70

Step 3) Sum of x-forces and sum of y-forces/Equilibrium equations

ƩF

_{x}= fs + F cos 70 + (-mg sin 70) = 0

ƩF

_{y}= n + (-F sin 70) + (-mg cos 70) = 0

ƩF

_{y}= n + (-F sin 70) + (-mg cos 70) = 0

ƩF

_{y}= n - F sin 70 - mg cos 70 = 0

ƩF

_{y}= n = F sin 70 + mg cos 70

ƩF

_{y}= n = 0.940F + (2.00)(9.8)cos 70

ƩF

_{y}=

**n = 0.940F + 6.704**

Step 4) Solve for static friction f

_{s}

f

_{s}= μ

_{s}n

f

_{s}= (0.300)n

f

_{s}= (0.300)(0.940F + 6.704)

**f**

_{s}= 0.282F + 2.0112Step 5) Find the HORIZONTAL force vector by plugging in f

_{s}in ƩF

_{x}

ƩF

_{x}= f

_{s}+ F cos 70 + (-mg sin 70) = 0

ƩF

_{x}= (0.282F + 2.0112) + 0.342F - 18.418 = 0

ƩF

_{x}= 0.282F + 0.342F + 2.0112 - 18.418 = 0

ƩF

_{x}= 0.624F - 16.4068 = 0

ƩF

_{x}= 0.624F = 16.4068

ƩF

_{x}=

**F = 26.29**N

(b) the normal force exerted by the incline on the block

n = 0.940F + 6.704

n = 0.940(26.29) + 6.704

**n = 31.4**N

MY QUESTION IS...

In STEP 2, why is the force "-mg sin 70" included in the X-DIRECTION and why is the force -mg cos 70 included in the Y-DIRECTION????

I thought the ONLY forces acting in the x-direction were static friction, f

_{s}and Fcos 70, while the only forces acting in the y direction were normal force, n, and -Fsin 70???

Help!