Aforce f = 94.9 n acts at an angle α = 35◦ with respect to the horizontal on a block of mass m = 29.3 kg, which is at rest on a horizontal plane. the acceleration of gravity is 9.81 m/s 2 . if the static frictional coefficient is µs = 0.69, what is the force of static friction? 29.3 kg µs = 0.69 94.9 n 35◦ 1. 144.69 n 2. 160.77 n 3. 37.5584 n 4. 53.6389 n 5. 198.329 n 6. 0 n 7. 235.887 n 8. 77.7375 n 9. 251.968 n

8) 77.7375 N

Explanation:

Given that

F = 94.9 N

α = 35◦

m = 29.3 kg

µs = 0.69

The component of force F

F1= 94.9 sin35◦ =54.43 N

F2= 94.9 cos35◦ =77.73 N

So

N + F1= mg

N= 29.3 x 9.81 - 54.43

N=233 N

So the maximum static friction force

fr = µs .N

fr = 0.69 x 233

fr = 160.77 N

The applied horizontal force = F2

F2=77.73 N

The applied horizontal force is less than the maximum value of static friction.It means that the block will not move.The block will remain in the static position.

So the static friction fr= f2

Static friction = 77.73 N

8. 77.7375

find the surface area of the sphere to the nearest square unit. use a calculator.

a. 254 in.2

b. 64 in.2

c. 127 in.2

d. 1,018 in.2