# Two forces on a point particle

1. Oct 24, 2008

### anteaters

1. The problem statement, all variables and given/known data

Besides its weight, a 4.20 kg object is subjected to one other constant force. The object starts from rest and in 1.20 s experiences a displacement of (3.80i hat - 3.30j hat) m, where the direction of j hat is the upward vertical direction. Determine the other force.

2. Relevant equations

F = ma; F =$$\sqrt{Fx2+Fy2}$$

3. The attempt at a solution

okay so i know that the weight is pulling down on the object at 41.16 N. So i need to find the direction and magnitude of a force countering it, so that the resultant vector will = 5.03, however i'm confused about what role 1.20 s has to do with the problem. any help?

2. Oct 24, 2008

### loonychune

You can break this problem down into its component constituents... the force due to gravity acts only in the j direction, so when this particle is displaced 3.80m in the i direction, in 1.20s, for which the acceleration is constant (because force is constant), you have:

$$3.80 = \frac{1}{2}a_x(1.20)^2$$

since the object starts at rest (this is a suvat equation). Then you have:

$$F_x = ma_x$$

Under the force of gravity alone, the object will travel some distance in j direction but has in fact travelled -3.30m in that direction... find out the difference and see what extra distance you need and then set up another equation the same as the one i have in x, with

$$s = 1/2a_y(1.20)^2$$

where s is this 'extra' distance.

The importance of the 1.20s then, is in calculating the component accelerations (and hence the forces) of the particle.

3. Oct 24, 2008

### anteaters

okay i did all that, and for the Fx, i still got the wrong answer. i got 23.76, but the program i'm using said i was within 10% of the answer. for the Fy, however, it said i was off by orders of magnitude. Here is what i did, maybe you can show me where i went wrong:

Yf = m(-g)t2 + s
then i plugged in numbers, solved for x, and put that back into that
s = 1/2ay(1.20)2,
found ay and then
plugged that into Fy = may
and got a final answer of 376.275, but it said i was off by orders of magnitude.