Defination of formula bar
2 answers:
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Hi there!
To find the appropriate force needed to keep the block moving at a constant speed, we must use the dynamic friction force since the block would be in motion.
Recall:
The normal force of an object on an inclined plane is equivalent to the vertical component of its weight vector. However, the horizontal force applied contains a vertical component that contributes to this normal force.
We can plug in the known values to solve for one part of the normal force:
N = (1)(9.8)(cos30) + F(.5) = 8.49 + .5F
Now, we can plug this into the equation for the dynamic friction force:
Fd= (0.2)(8.49 + .5F) = 1.697 N + .1F
For a block to move with constant speed, the summation of forces must be equivalent to 0 N.
If a HORIZONTAL force is applied to the block, its horizontal component must be EQUIVALENT to the friction force. (∑F = 0 N). Thus:
Fcosθ = 1.697 + .1F
Solve for F:
Fcos(30) - .1F = 1.697
F(cos(30) - .1) = 1.697
F = 2.216 N
Answer:
Using the pythagoras theorem
S²=9²+4²
S²=81+16
S²=97
S=9.85km.
In finding the direction
tan□=opposite/Adjacent
=4/9
□=23.96
¤=90-23.96
=66.03 degrees
9.85, N 66.03 E
