| T1 cos(35) - T2 |
 |
| m |
| 16.38 N - 12 N |
|
| 2.75 kg |
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. First choose a coordinate system --- +x = right and +y = up. Then show all the forces acting on the box.
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 the box accelerates it will accelerate horizontally.
Apply Newton's second law separately for the x- and y-directions. y-direction.
In the x direction, Σ Fx = m ax gives T1 cos(35) - T2 = m ax
| ax | = |
|
= |
|
= | 1.6 m/s2 |
The fact that the x-component of the acceleration is positive means the box accelerates to the right.
In the y direction, there is no acceleration, which means the forces have to balance.
Thus Σ Fy = m ay = 0 gives N + T1 sin(35) - mg = 0
Solving for N gives: N = mg - T1 sin(35) = 26.95 - 11.47 = 15.5 N
Note that the normal force is not equal in magnitude to the force of gravity.