Maths question

Hi, please could I have some help on resolving. I’ve drawn the forces out but for part b, I’m really confused because the two highlighted forces are not in the same direction and im trying to resolve vertically and horizontally?
Thanks!

Hi, please could I have some help on resolving. I’ve drawn the forces out but for part b, I’m really confused because the two highlighted forces are not in the same direction and im trying to resolve vertically and horizontally?
Thanks!

Check the direction of the frictional force - it acts to prevent slipping.
The reaction at the wall acts perpendicularly.
For (a) I would take moments about A to eliminate the forces acting at A

For b) as an alternative to resolving horizontally/vertically on the wall and ground (then pythagoras) you could simply draw the equillbrium force triangle with sides 9.56N, 20N and the balancing side (magnitude ground contact force) where the angle between the 9.56 and 20 sides is 55 degrees, so use the cos rule to get the unknown side.
Obviously, it helps to know/be able to do both approaches and theyre not that different.

I see, but how would I go about resolving the force at b to be vertical? I’ve tried drawing it so many different ways but it’s not properly vertical and horizontal

Draw a horizontal line from B (to the left), so begin the start of the reaction arrow, and drop a vertical from the end of the reaction arrow to form a right triangle where the legs are horizontal and vertical. A tiny bit of angle chasing gives the complementary angles of the right triangle.
The cos rule means you dont have to do this, so its worth practicing both in case one "doesnt work" in an exam. You simply add the weight (vertical downwards) to the end of the wall reaction and then for equilibrium, the missing force must be the missing side of the non right triangle (cos rule).

I don’t think I’ve drawn it right, I’ve followed what you said about going to the left and down but I don’t get how the green arrow is representing the reaction force at b, isn’t it always perpendicular to the wall?

The normal reaction, as the name suggests, is perpendicular to the wall. Thats the force arrow acting on AB at B. So your green arrow should be pointing to 11 o clockish. However you have the right idea for the red legs, if you correct the green arrow.
Its a smooth wall, so there is no component of the force at B parallel to the wall. It must be perpendicular to the wall.

Is this ok instead of drawing the red arrow down (because then I get a complete triangle)

No thats correct. I was being a bit loose with my language as the normal reaction points to the left (horizontal) and up (vertical) and you add vectors (legs) tip to tail. Then mark on the right angle for the force - wall and you should get the resolved force right triangle angles fairly easily.

Sorry yes I meant to draw friction the other way around. With the reaction at b, isn’t what I have drawn perpendicular to the grey line (the wall)?