# Question about resolving forces

## Homework Statement

On a model ship, the mast OC has length 50 cm and weight 20 newtons. The mast is hinged to the deck at O, so that it can rotate in the vertical plane of the ship. Small smooth rings are fixed at points A and B on the deck in this plane such that AO=OB=50 cm. Threads from C are passed through these rings, and held at their ends by two children who exert forces of P newtons and Q newtons respectively. If Q = 10, calculate the value of P needed to hold the mast in equilibrium at 40° to the horizontal deck.

2. The attempt at a solution

I tried finding a simultaneous equation by resolving the forces horizontally and vertically, but I was unable to do so as there are 6 different forces (P,Q,tension in AC,tension in BC, tension in OC, weight of OC) and the angles between P and the deck and Q and the deck are unknown (I found the angles between AC and the deck and CB and the deck as 20° and 70° respectively (or it could be the other way round?).).

## Answers and Replies

haruspex
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there are 6 different forces (P,Q,tension in AC,tension in BC, tension in OC, weight of OC)
Some of those are easily seen to be the same, if there is no friction in the rings.
This also makes some angles irrelevant.

• parsesnip
Some of those are easily seen to be the same, if there is no friction in the rings.
This also makes some angles irrelevant.
Ok. I think that |P| = |tension in AC| and |Q| = |tension in BC| (because the length of the strings has to stay equal). Then I get the equations:
Psinθ-Psin20°+Tsin40°-20-10sin70°+10sinφ=0
Pcosθ+Pcos20°-Τcos40°-10cos70°-10cosφ=0.

I think maybe Tsin40°=20?

haruspex
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I think that |P| = |tension in AC| and |Q| = |tension in BC|
Yes, but I would have said that it was because otherwise the rope would slide through the ring.
Then I get the equations:
Psinθ-Psin20°+Tsin40°-20-10sin70°+10sinφ=0
Pcosθ+Pcos20°-Τcos40°-10cos70°-10cosφ=0.

I think maybe Tsin40°=20?
What are θ and φ, and how do you get those equations?

θ is the angle between P and the horizontal and φ is the angle between Q and the horizontal.
I got those equations by resolving the forces vertically and horziontally (as the system is in equilibrium so net force = 0)

haruspex
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θ is the angle between P and the horizontal and φ is the angle between Q and the horizontal.
I got those equations by resolving the forces vertically and horziontally (as the system is in equilibrium so net force = 0)
You are ignoring the forces the rings exert on the ropes.
Just stick with the fact that the tension doesn't change as the rope passes through a ring and consider the forces on the mast.

• parsesnip
You are ignoring the forces the rings exert on the ropes.
Just stick with the fact that the tension doesn't change as the rope passes through a ring and consider the forces on the mast.

So the forces acting on the mast are the tensions of AC, AB and AO and the weight.
The equations are:
Pcos20=10cos70+Tcos40
Tsin40=20+10sin70+Psin20

P=72.3 N?

haruspex
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the tensions of AC, AB and AO
Did you mean that?
Pcos20=10cos70+Tcos40
What is T and how do you get this equation? What direction are you resolving in?
(Do not assume the force the deck exerts on the mast at O acts along the mast.)