Question

Two students are dragging a 25.0 kg object along the hall as shown in the diagram below. If the force of friction action on the object is 5.0 N, what is the acceleration of the object?
(Be sure to include direction)

Answer 1

so you can add your force vectors

Answer 2

F_net = mass * acceleration

F_net / mass = acceleration

Answer 3

I am in a confusion whether the object is traveling straight east or at a northeast direction because FT1 > FT2

Answer 4

actually the force of friction is opposite to the direction of movement

Answer 5

yes but the here the force of friction is less the force applied so it is moving forward

Answer 6

but what direction?

Answer 7

Wellfor one, you will be finding the force in both directions:\[\sum F_y\]\[\sum F_x\]

Answer 8

so first find the direction of movement, the opposite of that direction will be the friction vector

Answer 9

Then decide which way you want your positive direction to be before solving your equations.

Answer 10

Ft1 + Ft2
=< 40 cos 20 , 40 sin 20> + <-30 cos 25, -30 sin 25>
= < 40 cos 20 - 30 cos 25 , 40 sin 20 - 30 sin 25>

Answer 11

\[\sum F_y = ma_y = T_1-T_2\]

Answer 12

so I found the resultant vector of vectors Ft1 and Ft2 ,
the magnitude and direction you can find

Answer 13

Fnetx = FT1x + FT2x - Ffriction
= 37.59 + 21.65 - 5
= 54. 24

Answer 14

\[T_{1_y} = 40.0\sin (20^{\circ})\]\[T_{2_y} =25.0\sin(30.0^{\circ})\]

Answer 15

FT1y + FT2y = Fnety
(13.68) + (-12.5) = 1.18 N

Answer 16

My method is the same as @perl's except broken down into net directional form

Answer 17

So northeast direction?

Answer 18

Why are you addind both of the y directions? are they being pulled the same way?

Answer 19

Fnet is a total of all forces and since on of them is negative force

Answer 20

I added them