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Part A.)Six boxes held at rest against identical walls. Rank the boxes on the basis of the magnitude of the normal force acting on them. Rank from largest to smallest. To rank items as equivalent, overlap them.
1.)130N—>7kg
2.)150N—>1kg
3.)150N—>7kg
4.)120N—>3kg
5.)140N—>5kg
6.)140N—>3kg
(Since the boxes are at rest, Newton’s 2nd law dictates that the horizontal forces on each box must add up to zero. You can use this information to determine the normal forces. If two boxes are both pushed against the wall by the same force, then they should experience the same normal force.)
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Part B.) Rank the boxes on the basis of the frictional force acting on them.
Rank from largest to smallest. To rank items as equivalent, overlap them.
1.)130N—>7kg
2.)150N—>1kg
3.)150N—>7kg
4.)120N—>3kg
5.)140N—>5kg
6.)140N—>3kg
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Relevant knowledge
The normal force is the force that an object experiences when it comes into contact with a normal surface. Newton’s third law governs motion states that the normal force acting on an object is equal the force applied to it by the object normal the contact surface.
Newton’s Second Law, the concept that must answer the question, is Newton’s Second Law.
Applying Newton’s Second Law of Force, and the equilibrium condition of force, is the first step in ranking boxes based on frictional force and basis normal force.
The horizontal forces acting on the box later will be zero because it is now stationary. This gives the box its normal force.
Because the box is stationary, all vertical forces that act on it will be equal to zero. This is what gives the box’s frictional force its value.
According to Newton’s first law the net force that acts on a particle when it is at rest or moving with constant velocity within an inertial frame is zero. This is because it is equal to the vector sum of all forces. This is mathematically expressed as:
[katex]\sum {\vec F} = 0[/katex]
Here, [katex]F[/katex]
Newton’s second law says that the net force exerted on an object by a mass and its final acceleration is equal. The Newton’s second law equation is:
[katex]\sum {\vec F = m\vec a} [/katex]
Here, [katex]\sum {\vec F} [/katex]
The force is in equilibrium when it gives.
[katex]\Sigma F = 0[/katex]
Here, [katex]F[/katex]
(A)
Due to the surface supporting it, the normal force will be applied horizontally on the box.
The horizontal force acting upon the box will therefore be.
[katex]F – N = 0[/katex]
Here, [katex]F[/katex]
Rearrange the equation.
[katex]N = F[/katex]
[katex]150{\rm{ N}} > {\rm{140 N}} > 130{\rm{ N}} > 120{\rm{ N}}[/katex] is the descending order of the forces
The normal force on the box is equal to its normal force, so the ranking of normal forces on the box will be the same.
(B)
The net vertical force acting upon the box will equal zero.
[katex]F[/katex]0
The box’s weight will be a downward force that is directed towards the center of the earth, and the frictional force will oppose the direction the box moves.
The box will be affected by the vertical force.
[katex]F[/katex]1
Here, [katex]F[/katex]2
Rearrange the equation.
[katex]F[/katex]3
Here, [katex]F[/katex]4
The mass of each box in ascending order is [katex]F[/katex]5
This order will be used to rank boxes based on frictional forces. Part A – Ans
The rank of the normal force that acts on the boxes will [katex]F[/katex]6