When a 25-kg crate is pushed across a frictionless horizontal floor with a force of 200 N, directed 20 below the horizontal, th
e magnitude of the normal force of the floor on the crate isA. 25N
B. 68N
C. 180N
D. 250N
E. 310N
B. 68N
C. 180N
D. 250N
E. 310N
1 answer:
Answer:
Option E is correct 310N
Explanation:
Given that the force used to push the crate is F = 200N
The force directed 20° below the horizontal
Mass of crate is m = 25kg
Weight of the crate can be determine using
W = mg
g is gravitational constant =9.8m/s²
W = 25×9.8
W = 245 N
Check attachment. For free body diagram and better understanding
Using newton second law along the vertical axis since we want to find the normal force
ΣFy = m•ay
ay = 0, since the body is not moving in the vertical or y direction
N—W—F•Sin20 = 0
N = W+F•Sin20
N = 245+ 200Sin20
N = 245 + 68.4
N = 313.4 N
The normal force is approximately 310 N to the nearest ten
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Answer:
6.19 x m/
Explanation:
For this exercise we need to sum the forces on the y-axis and x-axis as follows:
∑ = N - mg = m.
= 0
From the exercise, we deduce there is no motion in y-axis, so:
N = mg
Then for x-axis we have:
∑ = H -
= m.
= 0
Now, from the exercise we deduce that we are looking for the greatest static friction which means to have the maximun static friction to start moving, so at this point the acceleration is zero, so we can find horizontal force (H), which then will act in the airplane to move it. Therefore we have:
H = =
=
N =
mg
H = (0.76)(84Kg)(9.8m/)
H = 625.63 N
Now we apply this force to the weight of the plane to find the greatest acceleration the mann can give to start moving the plane.
a = =
a =
a = 6.19 x m/