Goal: Problem solving with dynamics
Source: UMPERG-ctqpe21
Two blocks rest on a frictionless surface. Both blocks move to the right
with acceleration of 2 m/s2. The force on the big block due
to the small block is
Goal: Reasoning with Coulomb’s law
Source: UMPERG-A2LEM6
The diagrams show two uniformly charged spheres. The charge on the
right sphere is three times as large as the charge on the left sphere.
The arrows on each charge represent the force on the charge. Which
force diagram best represents the magnitude and direction of the
electric forces on the two spheres?
(4) By Newton’s third law the forces are equal and opposite.
Goal: Reasoning with electric forces
Source: UMPERG-A2LEM5
What is the direction of the electrical force on the charge q shown in
the diagram?
(3)
Goal: Problem solving with Coulomb’s law
Source: UMPERG-A2LEM4
What is the size of the electrical force on the charge q shown in the diagram?
(3) It is a common mistake to forget to take the y-component of the
force.
Goal: Reasoning with Coulomb’s Law
Source: UMPERG-A2LEM3
Three charges are positioned on the x-axis as shown. What is the
direction of the electrical force on the right-most charge?
(1) Students who forget to square the denominator in Coulomb’s law will answer (5).
Goal: Hone the vector nature of electric force.
Source: UMPERG-A2LEM2
Two charges, a negative charge Q, and a positive charge q, are
positioned as shown in the diagram. What is the direction of the
electrical force on q due to Q?
(3) The force is attractive.
Goal: Hone the vector nature of electric force.
Source: UMPERG-A2LEM1
Two positive charges, Q and q, are positioned as shown in the diagram.
What is the direction of the electrical force on q due to Q?
(1) The force is repulsive.
Goal: Reasoning about forces and torques.
Source: UMPERG-ctqpe158
A uniform rod is hinged to a wall and held at a 30° angle by a thin
string that is attached to the ceiling and makes a 90° angle to rod.
Which statement must be true?
(4) This is easily determined by considering torques about the hinge.
The hinge force cannot be purely vertical because there is a horizontal
component to the tension that must be balanced. Also the hinge force
cannot be purely horizontal or the rod would rotate counterclockwise
about its center. Since the hinge force must have a horizontal component
in the opposite direction as the horizontal component of the tension,
(3) cannot be true either.
Goal: Problem solving
Source: UMPERG-ctqpe166
A uniform disk with mass M and radius R sits at rest on an incline
30° to the horizontal. A string is wound around disk and attached
to top of incline as shown. The string is parallel to incline. The
friction force acting at the contact point is:
(3) Balancing torques about the center of the disk determines that the
friction force points up and is equal to the tension in the string. (The
other forces, gravity and normal pass through the point and contribute
no torques.) Balancing torques about the contact point determines the
tension readily.
Goal: Problem Solving
Source: UMPERG-ctqpe164
A uniform disk with mass M and radius R sits at rest on an incline
30° to the horizontal. String is wound around disk and attached to
top of incline as shown. The string is parallel to incline. The
tension in the string is :
(4) This problem can be solved a variety of ways. The simplest method is
to balance torques about the contact point. This situation is an
excellent one for discussing the advantages of thinking about preferred
points about which to write the rotational dynamics equation.
Commentary:
Answer
(7) The force on the small block must cause the specified acceleration.
The 3rd law requires that the force on the big block be equal and
opposite. The magnitude is 4N but it is directed to the left.