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:
The weight lifter would not get past this sticking point.
Explanation:
Generally torque applied on the weight is mathematically represented as
T = F z
To obtain Elbow torque we substitute 4000 N for F (the force ) and 2cm for z the perpendicular distance
So Elbow Torque is
To obtain the torque required we substitute 300 N for F and 30cm
So the Required Torque is
Now since it mean that the weight lifter would not get past this sticking point
See the attached picture for answers
