# Resolving force

1. Jul 15, 2011

### drawar

1. The problem statement, all variables and given/known data
Particle P has a mass of 1kg and is confined to move along the smooth vertical slot due to the rotation of the arm Ab. Assume that at the instant as shown, the acceleration of the particle P is 2.46 m/s^2 upward. Determine the normal force N(rod) on the particle by the rod and the normal force N(slot) on the particle by the slot in the position shown. Frictions forces are negligible.

----N(rod) ---- N(slot)
A. 14.2 N ---- 7.1 N
B. 10.5 N ---- 3.6 N
C. 7.1 N ---- 14.2 N
D. 3.2 N ---- 1.6 N

2. Relevant equations F= ma

3. The attempt at a solution

F = ma = 2.46 N
The force acting on the particle can be resolved into 2 components
N(rod) = F x cos(30) = 2.13 N
N(slot) = F x sin(30) = 1.23 N

The answer does not match any of the choices provided. Can someone help me figure out what I'm doing wrong? Thanks

Last edited: Jul 16, 2011
2. Jul 15, 2011

### cupid.callin

Hi drawar
Welcome to PF!

a diagram would be nice.

3. Jul 16, 2011

### drawar

Hi cupid.callin :)
I've already drawn a vector diagram showing my work

#### Attached Files:

• ###### image1.png
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4. Jul 16, 2011

### RTW69

Your N(slot) force is in the wrong direction. N(rod) is correct. If you resolve N(rod) onto forces in the x and y directions you will see that N(slot) must act in the positive x-direction (to the right). Draw a free body diagram with all forces. Identify direction of acceleration. Use F=ma for x and y directions.

5. Jul 16, 2011

### darkxponent

let AP= l
and vertical position of particle = y
now: y= l*sine(theeta)

velocity of particle= v= dy/dt= l*cos(theeta)*d(theeta)/dt
v = Vr*cos(theeta)

acc of paticle = a= dv/dt = (-)*l* sine(theeta) *w + l*cos(theeta)* q {q=angular acc of rad, w= angular vel of rad}

acc of particle = -Vr*sin(theeta) + Ar*cos(theeta)

{Vr , Ar = velocity of rod, Accc of rod at particles point}

this means :: DATA insufficient...either posotion or velocity particle is required

6. Jul 16, 2011

### Staff: Mentor

How is the gravitational force being accounted for (the particle has a mass, therefore a weight)? I don't see it in your FBD or calculations.

7. Jul 16, 2011

### cupid.callin

Am i the only one who still cant see the pic?

8. Jul 16, 2011

### drawar

Thanks for all your kind help :-)
I totally forgot about the gravitational force. I've just give it another try and here is the result, hope someone might have a check at this:

Resolving vertically upward:
$$N_{rod} \times \cos 30^o - mg = ma \Rightarrow N_{rod} = \frac{{m(g + a)}}{{\cos 30^o }} = 14.2$$(1 d.p)

Resolving horizontally rightward:
$$N_{slot} - N_{rod} \times \sin 30^o = 0 \Rightarrow N_{slot} = \frac{{N_{rod} }}{{\sin 30^o }} = 7.1$$

Hence choose A

edit: @cupid.callin: I'm afraid so. Seems like your browser is having trouble reading image from imgur.com. I'll upload it to another host

Last edited: Jul 16, 2011