A box of mass 2.75 kg sits on a table. Neglect friction. You pull on a string tied to the right side of the box, exerting a force of 20.0 N at an angle of 35.0 degrees above the horizontal. Your friend exerts a horizontal force of 12.0 N by pulling on a string on the other side of the box.
(a) What is the box's acceleration?
(b) What is the normal force exerted on the box by the table?
Draw the free-body diagram - show all the forces acting on the box.
Think about what the box will do.
The force of gravity is mg = 2.75 x 9.8 = 26.95 N, directed down. The vertical component of your force is less than that, so the box remains on the table. If it accelerates it will accelerate horizontally.
Choose a coordinate system.
Assume the box accelerates to the right, so +x = right and +y = up.
Apply Newton's second law twice, once for the x-direction and once for the y-direction.
In the x direction, summing the forces gives:
Σ Fx = m ax
T1 cos(35) - T2 = m ax
ax | = |
|
= |
|
= | 1.6 m/s2 |
The fact that it's positive means the box does accelerate to the right.
In the y direction, there is no acceleration, which means the forces have to balance.
Σ Fy = m ay = 0
N + T1 sin(35) - mg = 0
N = mg - T1 sin(35)
N = 26.95 - 11.47 = 15.5 N
Note that in this case the normal force is not equal in magnitude to the force of gravity.