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Problem: Box pulled at an angle over a horizontal surface

A box of 5.0 kg stays on a frictionless horizontal surface.

Bob then starts pulling the box with a force of 100 N at an angle of 15° with the horizontal.

  • Draw a free-body diagram for the box.
  • What is the resultant force acting on the box?
  • What is the acceleration of the box?
  • What is the normal force that the surface exerts on the box?

Solving the problem

Before we can solve this problem, we need to understand what is happening. Let's try to draw a sketch, using the information that was specified in the problem description.

So, we probably need to draw a horizontal surface, a rectangular box sitting on the surface, and indicate that the box is being pulled by a force that makes 15° with the horizontal:

Sketch of Bob pulling the box with a force which makes 15 degrees with the horizontal15°Pull

The next step is to look carefully at our sketch, and try to enumerate all the forces that are acting on the box.

There are clearly 3 forces acting:

  • Bob's pulling force, which we'll indicate with F
  • the gravitational force mg
  • and the normal force N, which the surface exerts to prevent the penetration of the box

At this point we are ready to draw a free-body diagram for the box:

Free-body diagram of boxmgNF15°

Now, let's carefully look at what we know, and what the problem is asking:

We know the mass of the box (5.0 kg) and Bob's pull (100 N at 15° angle with the horizontal).

The problem is asking us to draw a free-body diagram for the box (which we already did), then to determine the resultant force that acts on the box, the acceleration that the box has as a result, and the normal force that the surface exerts on the box.

In short:

We know

m = 5.0 kg
F = 100 N
θ = 15°

We want to know

R = ?
a = ?
N = ?

We can start by finding the resultant force.

So, let's draw the coordinate axes on our diagram, with the x axis in the direction of motion, and find the x and y components of the 3 forces that act on the box:

Free-body diagram with coordinate axesxy15°mgNFxFyF
Fx = F cos 15°
Nx = 0
mgx = 0
Fy = F sin 15°
Ny = N
mgy = −mg

We can now find the x and y components of the resultant force:

x:

Rx = Fx + Nx + mgx
Rx = F cos 15° + 0 + 0
Rx = F cos 15° (1)

y:

Ry = Fy + Ny + mgy
Ry = F sin 15° + N + (−mg)
Ry = F sin 15° + Nmg (2)

Since the motion is along a horizontal surface, the y component of the resultant force, Ry, must be zero. Otherwise there would be an acceleration in the y direction (for Newton's 2nd Law), which is not the case.

Ry = 0

Therefore, the magnitude of R will be equal to the magnitude of its x component.

The x component, Rx, is positive because the box is accelerating in the positive x direction:

Rx > 0
R = Rx

We have already determined Rx in Eq. (1):

Rx = F cos 15°
R = Rx = F cos 15°
R = (100 N) (cos 15°)
R = 97 N

And R will have direction in the positive x axis:

The resultant force which acts on the boxxyR

Next, let's find the acceleration of the box.

We know the mass of the box, and we just found the resultant force acting on the box. We can apply Newton's 2nd Law to get the acceleration:

R = ma
aR
m

The direction of a is the same as that of R. Let's calculate the magnitude of a:

aR
m
a97 N
5.0 kg
a = 19 N/kg = 19 m/s2

The last unknown on our list is the normal force N.

N has direction in the positive y axis. But what about the magnitude?

We saw N appear in Eq. (2):

Ry = F sin 15° + Nmg

And we know that Ry is zero, therefore:

0 = F sin 15° + Nmg

This is an equation with one unknown (N). So, let's solve it:

0 = F sin 15° + Nmg
N = F sin 15°mg
N = −F sin 15° + mg
N = mgF sin 15°
N = (5.0 kg) (9.8 N/kg) − (100 N) (sin 15°)
N = 49 N26 N
N = 23 N

We have found all the unknowns the problem asked us:

R = 97 N
a = 19 m/s2
N = 23 N

Tips & Tricks

  • Don't make the mistake to assume that the magnitude of the normal force is always equal to the magnitude of the gravitational force mg. As we saw in this problem, the normal force had magnitude equal to mg − F sin 15°.
  • Since the definition of newton is: 1 N = 1 kg · m/s2, it follows that: 1 N/kg = 1 m/s2. This means that we can use N/kg and m/s2 interchangeably to express accelerations.
  • If an object moves horizontally, the vertical component of the resultant force is zero.

Exercises

#1

A crate of mass 31.0 kg is pulled over the floor with a force of 275 N. The force makes an angle of 24.0° with the horizontal. Assuming there is no friction between the crate and the floor, find the resultant force acting on the crate, the acceleration of the crate, and the magnitude of the normal force.

Solution

R = 251 N
a = 8.10 m/s2
N = 192 N
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