Moment double pivot - What are the normal contact forces at P and Q?

Answer: Option A

Firstly, you need to know that at both pivot P and Q, there are normal contact forces acting upwards on the plank.

2 conditions for the plank to be in equilibrium:

1) Sum of clockwise moment = Sum of anticlockwise moment (POM)

2) Net force = 0N (i.e. total upward forces = total downward forces)

To find the value of the normal force, you can choose to take pivot about either P or Q. If you take moment about P, the normal force at P will not be in the Principle of Moments equation Ma = Mc as the line of action of the normal contact forces at P passes through the pivot P, there is no perperdicular distance, hence no moment created. We can determine the value of the normal contact forces at Q.

Once normal force at Q is found, you can use the concept net force = 0N to find the normal force at P straight away, which is faster.

https://youtu.be/d8q0hYielsQ

What is the tension in the string connecting block A and B?

Answer: Option C

For such questions, always look at the blocks as one single system and find the acceleration of the whole as one. That acceleration is the same for both blocks.

Then apply free body diagram (the one with the unknown T and with the least number of forces acting), find the unknown T.

https://youtu.be/sZ8b46rIwd0

Refer to this post for more examples of such questions (some involve friction)

https://www.sgphysicstuition.com/post/finding-forces-on-multiple-bodies

Which speed-time graph represents the resultant-time graph?

Answer: Option D

From the resultant force-time graph, from time to time 0 to t, there is a decreasing resultant force. Using Fnet = ma, as the Fnet decreases, the acceleration is decreasing. But note that decreasing acceleration does not mean speed is decreasing. It means the speed is still increasing, just that the increment is getting smaller per unit time.

After time t, there is not resultant force, hence no acceleration. Since the object is already moving, it will continue to move at constant speed in a straight line.

On the speed time graph, D fits the scenario.

https://youtu.be/5BPXP3c_d_Q?si=Lfqnz90OKaLAnT1k