Homework Help: Finding angle for specific moment

1. Oct 14, 2015

werson tan

1. The problem statement, all variables and given/known data
i have question with part B
my working is
13.2 cos tetha (0.86x10^-3)+ 13.2 sin tetha (1.22x10^-3) = 1.95
i gt stucked here. how to proceed? i only have one equation , how to solve two unknown ?

2. Relevant equations

3. The attempt at a solution

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2. Oct 14, 2015

andrevdh

You should not assume the moment is 1.95 for part A. From the given information you should be able to see that theta is 30o, which enables you to calculate the moment of P about A for question a.

3. Oct 14, 2015

werson tan

i have done part A , i just have problem with part b , i did not assume the moment is 1.95 for part A. i gt -1.788N for part a

4. Oct 14, 2015

andrevdh

Another way is to use the formula τ = P r sin(θ) where θ is the angle between P and r. You need to make a drawing to figure our what the relationship between θ and α is then.

5. Oct 14, 2015

werson tan

(13.2cos (alpha) ) ( surd(0.086² +0.122² ) )= 1.95 , alpha = 8.23 degree , still wrong !

6. Oct 14, 2015

andrevdh

Did you take the square root to obtain r? The angle in the equation is not necessarily alpha. Make a drawing to discover the relationship.

7. Oct 30, 2015

werson tan

my working is 13.2 (1-(sin theta)^2) )^(0.5) (0.86x10^-3) + 13.2(sin theta)(122x10^-3) = 1.95 , so i gt the theta = 56.42 , but the ans given is 50.6 degree and 59.1 degree

8. Oct 30, 2015

andrevdh

I seem to get 47.5o for α

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9. Oct 30, 2015

haruspex

You do not have two unknowns. You have two functions, cos and sin, of one unknown.
Do you know the formula for sin(a+b)? Can you see how to apply it in reverse to an expression like $A\cos(\theta)+B\sin(\theta)$?

10. Oct 30, 2015

haruspex

Unfortunately the arc sin function gets very steep near 1, so it's hard to get enough precision. I get more like 46.6 or 63 degrees.

11. Oct 30, 2015

werson tan

can i do in this way ?
13.2 (1-(sin alpha)^2) )^(0.5) (0.86x10^-3) + 13.2(sin alpha)(122x10^-3) = 1.95 , so i gt the alpha = 56.42 , but the ans given is 50.6 degree and 59.1 degree

12. Oct 30, 2015

haruspex

All those numbers make it too hard to follow what you are doing. Use symbols and show your working to get that equation.

13. Oct 30, 2015

werson tan

sorry , pls ignore my previous working. i have changed the cos aplha = (1- (sin aplha )^2 )^0.5 , but i still coulnt get the ans , since the sin alpha value is greater than 1

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14. Oct 30, 2015

haruspex

$\sqrt{1-\sin^2(\alpha)}$ is not $1-\sin(\alpha)$.