Vector Analysis and Forces Acting on an Object - 1 - Free MCQ Practice

• Normal force is the force acting on an object in a direction perpendicular to the surface of contact.
• It is the force that keeps objects from sinking into the surface and must be large enough to negate any acceleration into the surface.
• When an object is traveling at a constant velocity, it is in equilibrium, which means that it experiences no net force. The only force, other than the normal force, acting on the block is the gravitational force. A free-body diagram will show that the gravitational force will be equal in magnitude and in the opposite direction to the normal force.
• Gravitational force for objects close to the surface of the earth is F_g = mg.
• The normal force will be large enough in magnitude to yield zero net force.
  F_net = mg - F_N
  0 = 10010 - F_N
  F_N = 1000 N

• Drawing a free-body diagram to show all the forces acting on the rock will help us set up the equation to obtain the net force.
• The forces acting on the rock are the gravitational force, the normal force, the force of the man pushing the rock, and the frictional force.
• The gravitational force and the normal force negate each other in the vertical direction; therefore the net force will be in the horizontal direction.
• In the horizontal direction, the force of the push works against the force of friction. Since the rock moves, the force of the man’s push overcomes the friction in the direction of the net force.
• The frictional force from kinetic friction is defined as the product of the normal force and the coefficient of kinetic friction. We can use all of this information to substitute for values in finding the net force on the rock.
  

• Since the normal force will have a magnitude of whatever it takes to keep an object from penetrating the surface, drawing a free-body diagram indicating all the forces acting on box A will help determine the value of the normal force.
• Drawing the free-body diagram of box A, we notice that there is the upwards normal force exerted by the ground. In the downward direction, there is the gravitational force of box A as well as the applied weight of boxes B and C pushing down.
• Since box A is in equilibrium, the sum of forces on box A will equal zero. This means that:
  

Drawing a free-body diagram will indicate that the refrigerator does not have net forces and is in equilibrium.
The maximum force of static friction that the floor could exert on the refrigerator is
F_{maximumstaticfriction}= μ_s mg
​F_{maximumstaticfriction} = 0.4100 9.8
F_{maximumstaticfriction} = 400 N (approximately)
However, static friction is only as large as necessary for the object to stay static and counteract the opposing force until the force due to static friction reaches its maximum amount.
The force on the refrigerator opposing static force in this scenario is the child’s push which amounts to only 50 N, much less than the 400 N threshold.

• Drawing a free-body diagram of a elevator in motion will elucidate the forces on the elevator.
• The forces on the elevator when it is moving downwards at a constant velocity are the gravitational force and the force due to the tension on the cable.
• If the elevator is moving at constant velocity, then there is no net acceleration and no net force. The tension in the cable must equal the gravitational force of the elevator.

• Drawing a free-body diagram of the crate in motion will elucidate the forces on the crate.
• The forces present in the vertical direction are the normal force and the gravitational force; these forces cancel out. The forces present in the horizontal direction are the force due to kinetic friction and another applied force in the opposite direction of the friction since there is a constant velocity that does not dissipate due to the friction.
• Since there is a constant velocity, there is no net acceleration and no net force.

• Drawing a free-body diagram of the coin will elucidate the forces on the coin.
• The forces included in this problem are the gravitational force of the coin, the force of the push applied by Jennifer on the coin, and the normal force applied the surface of the table on the coin.
• The normal force is the force that keeps objects from sinking into the surface and must be large enough to negate any acceleration into the surface.
• The force from Jennifer and the gravitational force pushes down on the coin, while the normal force from the surface of the table pushes up on the coin.
• Since the coin is in static equilibrium, the sum of forces will equal to 0. Converting to kg, the coin has a mass of 0.005 kg.
  

• The magnitude of the normal force does not have to equal its gravitational force. It is possible for there to be an additional force acting in the opposite direction of the normal force that will cause an increase of the magnitude of the normal force. It is also possible for the normal force to be equal to a directional component of the gravitational force, such as in the case where the object is elevated on an incline.
• An object can be in motion experiencing constant velocity. Therefore there is no requirement for the object to be at rest.
• When all the forces acting upon an object balance each other, the object will be at equilibrium, and will not accelerate. There is no requirement for the forces to be equal in magnitude in all of its sides. There is a requirement for the sum of forces on each side to balance the sum of forces on its opposite side.

• Drawing a free-body diagram of the passenger will elucidate the forces on the passenger.
• The passenger experiences a gravitational force down the elevator while experiencing a normal force up in the upwards direction from the elevator floor surface.
• The net acceleration of the passenger is the same as the elevator, which is 3 m/s² in the upwards direction.
• The net force of the passenger is the product of the mass of the passenger and the net acceleration.
• The net force of the passenger should equal the sum of individual forces on the passenger. Individual forces that are directed upwards towards the direction of the net force are positive in sum of forces equation:
  F_net = F_normal - F_{gravitational}
  ma_net = F_normal - mg
  F_normal = mg + ma_net
  F_normal = 50 × 10 + 50 × 3
  F_normal = 500 + 150
  F_normal = 650 N

• To figure out the distance traveled after the car stops, we need to find the acceleration of the car after it stops its wheels from moving. Drawing a free-body diagram of the car will assist us in elucidating the net force and net acceleration experienced by the car.
• A free-body diagram of the car moments after it stops its wheels from moving demonstrates that it experiences a gravitational force downwards, a normal force upwards, and a force of friction against its direction of movement, dissipating its movement and making it reach its stopping point.
• Since the normal and gravitational forces cancel out, the force due to kinetic friction is the only force relevant to our determination of the net acceleration. This force is represented by
  F_net  = μ_k
  ma_net = μ_k
  ​a_net = μk
• Given the acceleration, we can use the kinematics formula v_final² -v_initial² = 2ad to solve for the distance traveled: