Answer:
a) -4 N
b) +4 N
Explanation:
Draw a free body diagram for each block.
For the large block, there are 2 forces: 12 N pushing to the right, and F pushing to the left.
For the small block, there is 1 force, F pushing to the right.
There are also weight and normal forces in the vertical direction, but we can ignore those.
Sum of forces on the large block in the x direction:
∑F = ma
12 − F = 4a
Sum of forces on the small block in the x direction:
∑F = ma
F = 2a
2F = 4a
Substitute:
12 − F = 2F
12 = 3F
F = 4
The small block pushes on the large block 4 N to the left (-4 N).
The large block pushes on the small block 4 N to the right (+4 N).
Explanation:
Archimedes' principle states that the upward buoyant force which is exerted on body when immersed whether fully submerged or partially in the fluid is equal to weight of fluid which body displaces and this force acts in upward direction at center of mass of displaced fluid.
Thus,
<u>Weight of the displaced fluid = Weight of the object - Weight of object in fluid.</u>
Answer:
The force pulling the roller along the ground is 128.55 N
Explanation:
A force of 200 N acting at an angle of 50° with the ground level
This force is pulled a garden roller
We need to find the force pulling the roller along the ground
The force that pulling the roller along the ground is the horizontal
component of the force acting
→ The force acting is 200 N at direction 50° with ground (horizontal)
→ The horizontal component = F cosФ
→ F = 200 N , Ф = 50
→ The horizontal component = 200 cos(50) = 128.55 N
128.55 N is the horizontal component of the force that pulling the
roller along the ground
<em>The force pulling the roller along the ground is 128.55 N</em>
