I kg 12 N 30° Figure 1 ll by a 12 N force as 1. A 1.0 kg wood block, initially a
ID: 1872972 • Letter: I
Question
I kg 12 N 30° Figure 1 ll by a 12 N force as 1. A 1.0 kg wood block, initially at rest, is pressed against a vertical wood wa shown in Figure 1. The coefficient of static friction (ps) for wood on wood is 0.5. Draw the FBD for the wood block with a coordinate system. Write out Newton's 2nd law for this block. What is the direction of the friction? Why? Hint: What direction would the block move initially without friction? a. b. Calculate the acceleration of the block. Will the block move upward, move downward, or stay at rest? c. d.Explanation / Answer
given mass of wooden block = m = 1 kg
coefficient of static friciotn k = 0.5
F = 12 N
theta = 30 deg
a. the FBD is as under
b. hence from force balance
Fcos(30) + f = mg
Fsin(30) = N
f = 9.81 - 12*cos(30) = -0.58230484
hence as f < 0
hence direcitn of friciton is downwards
c. now, acceleration of the block is a
then
Fcos(30) + f - mg = ma
then
12cos(30) + 0.5*12*sin(30) - 9.81 = a
a = 3.5823 m/s/s
d. the block will move upwards