A crane has an arm to which is attached a frictionless pulley. A cable passes over the pulley and supports a load of 10 kN as shown.

Question 12

A crane has an arm to which is attached a frictionless pulley. A cable passes over the pulley and supports a load of 10 kN as shown.

The crane arm exerts a force F on the pulley.

What is the value of F ?

A 5.0 kN                      B 8.7 kN                      C 10 kN                       D 17 kN

Reference: Past Exam Paper – March 2016 Paper 12 Q2

Solution:

Answer: D.

The force of 10 kN in the cable and the load of 10 kN both have a component along the crane arm.

The force F exerted by the crane arm on the pulley is the sum of these 2 components.

Component of force in cable along the crane arm = 10 cos 30°

Component of load along the crane arm = 10 cos 30°

Considering the forces on the pulley,

F = 10 cos 30° + 10 cos 30°

Force F = 17.32 kN = 17 kN

There is a current from P to R in the resistor network shown. The potential difference (p.d.) between P and Q is 3 V.

Question 32

There is a current from P to R in the resistor network shown.

The potential difference (p.d.) between P and Q is 3 V.

The p.d. between Q and R is 6 V.

The p.d. between P and S is 5 V.

Which row in the table is correct?

Reference: Past Exam Paper – June 2016 Paper 11 Q35

Solution:

Answer: A.

Current flows from P to R.

Recall that current cannot flow in opposite directions in the same wire.

There are different paths that the current could possibly flow.

Path 1: P to Q to R

Path 2: P to Q to S to R

Path 3: P to S to R

Path 4: P to S to Q to R

The p.d. between P and R is constant. This means that the sum of p.d. in any of the paths should be the same.

Let the p.d. between Q and S be QS and the p.d. between S and R be SR.

Considering the different paths,

Path 1: p.d between P and R = 3 + 6 = 9 V

Path 2: p.d. between P and R = 3 + QS + SR

The sum of p.d. in path 2 should also be 9 V as the p.d. between P and R is constant.

3 + QS + SR = 9

QS + SR = 9 – 3 = 6

QS + SR = 6               eq (1)

Path 3: p.d. between P and R = 5 + SR = 9

SR = 9 – 5 = 4 V

Substituting SR = 4 V in eq (1),

QS + 4 = 6

QS = 6 – 4 = 2 V

Path 4 may not be a possible path as the sum of p.d. is different.

A uniform diving-board is held by two fixed rods at points P and Q. A person stands at end R of the diving-board, as shown.

Question 32

A uniform diving-board is held by two fixed rods at points P and Q. A person stands at end R of the diving-board, as shown.

The forces exerted by the rods on the board are vertical. The board remains in equilibrium as the person slowly moves towards point Q from end R.

Which row describes the changes to the forces exerted by the rods on the board?

force at P                    force at Q

A         decreases                   decreases

B         decreases                   increases

C         increases                     decreases

D         increases                     increases

Reference: Past Exam Paper – June 2018 Paper 12 Q13

Solution:

Answer: A. 

For the board to remain in equilibrium, the resultant moment on it should be zero. That is, clockwise moment = anticlockwise moment.

The person’s weight acts downwards, causing a clockwise moment about the rods.

The rod at P exerts a downward force while the rod at Q exerts an upward force. Both of the forces causes a moment (in opposite direction) which balance the moment caused by the weight of the person.

We may take the pivot to be either at P or Q. When the pivot is taken to be at P, then the moment caused by the force at P is zero (as the perpendicular distance is zero). Similarly if the pivot is taken at Q, then the moment due to the force at Q is zero.

If the pivot is at P, the upward force at Q causes an anticlockwise moment. In the same way if the pivot is at Q, the downward force at P again causes an anticlockwise moment.

As the person moves towards the rods, the moment of the person’s weight at both P and Q decreases, and therefore the forces exerted by the rods at both P and Q must also decrease.