Explanation:
The rod is uniform, so the center of gravity is at the center, or 0.75 m from the end. The wedge is 0.5 m from the end, so the center is 0.25 m from the wedge.
Sum the torques about the wedge (it may help to draw a diagram first). Take counterclockwise to be positive.
∑τ = Iα
W (0.25 m) − (100 N) (0.50 m) = 0
W = 200 N
Sum the forces in the y direction.
∑F = ma
F − 100 N − 200 N = 0
F = 300 N
Answer:
( 1000 × 4 = 4,000) (800×3= 2400) (800×2=1600) the answer is 1600 hope it helps
Answer:
Explanation:
The question relates to motion on a circular path .
Let the radius of the circular path be R .
The centripetal force for circular motion is provided by frictional force
frictional force is equal to μmg , where μ is coefficient of friction and mg is weight
Equating cenrtipetal force and frictionl force in the case of car A
mv² / R = μmg
R = v² /μg
= 26.8 x 26.8 / .335 x 9.8
= 218.77 m
In case of moton of car B
mv² / R = μmg
v² = μRg
= .683 x 218.77x 9.8
= 1464.35
v = 38.26 m /s .
The answer is A. <span>Some work input is used to overcome friction. </span>
