The Object Sliding Horizontally calculator analyzes a block on a level surface when an external force is applied at an angle. The model includes the applied force, gravity, the normal force, kinetic friction, and an optional constant drag force. The goal is to find the horizontal net force and then use Newton’s second law to compute the block’s acceleration.
1. Coordinate convention
The calculator uses the usual convention:
- positive \(x\) is to the right,
- positive \(y\) is upward,
- the applied-force angle \(\theta\) is measured from the positive horizontal axis,
- the surface is horizontal, so vertical acceleration is assumed to be zero while contact is maintained.
2. Resolving the applied force
An applied force \(F\) at angle \(\theta\) is separated into horizontal and vertical components:
\[
\begin{aligned}
F_x &= F\cos\theta \\
F_y &= F\sin\theta
\end{aligned}
\]
If the force is pulled upward, \(F_y\) is positive and reduces the normal force. If the force is pushed downward, \(F_y\) is negative and increases the normal force.
3. Weight and normal force
The true weight of the block is:
\[
W = mg
\]
Since the block is assumed to remain on a horizontal surface, the vertical forces balance:
\[
N + F_y - mg = 0
\]
Solving for the normal force gives:
\[
N = mg - F_y
\]
This is why pulling upward reduces friction: it reduces the normal force. Pushing downward increases friction because it increases the normal force.
4. Loss of contact
If the upward component of the applied force is larger than the weight, then the formula gives \(N<0\). A surface cannot pull downward on the block, so a negative normal force is not physically possible for contact. In a physical contact model:
\[
N_{\text{physical}} = \max(0, mg-F_y)
\]
If \(N=0\), the block has lost contact with the surface and kinetic friction becomes zero. The ordinary horizontal sliding model then becomes incomplete because the object may also accelerate vertically.
5. Kinetic friction
Kinetic friction is modeled as:
\[
f_k = \mu_k N
\]
The direction of kinetic friction is opposite the sliding direction. If the block is moving right, friction points left. If the block is moving left, friction points right.
In component form:
\[
F_{\text{friction},x} = -s f_k
\]
where \(s=+1\) for motion to the right and \(s=-1\) for motion to the left.
6. Constant drag
The calculator treats drag as a constant force \(D\) opposing the selected direction of motion:
\[
F_{\text{drag},x} = -sD
\]
This is a simplified model. Real air resistance often depends on speed, but a constant drag term is useful for introductory force-balance problems.
7. Net horizontal force
The horizontal net force is the sum of the horizontal components:
\[
\sum F_x = F_x + F_{\text{friction},x} + F_{\text{drag},x}
\]
For a block moving right, this becomes:
\[
\sum F_x = F\cos\theta - \mu_kN - D
\]
Substituting \(N=mg-F\sin\theta\):
\[
\sum F_x = F\cos\theta - \mu_k(mg-F\sin\theta)-D
\]
8. Acceleration
Newton’s second law in the horizontal direction gives:
\[
\sum F_x = ma_x
\]
Therefore:
\[
a_x = \frac{\sum F_x}{m}
\]
A positive result means acceleration points to the right. A negative result means acceleration points to the left. If the object is already moving right and the acceleration is negative, the object is slowing down.
9. Worked example
Suppose:
- \(m = 5\,\mathrm{kg}\)
- \(F = 20\,\mathrm{N}\)
- \(\theta = 30^\circ\)
- \(\mu_k = 0.20\)
- \(D = 1.5\,\mathrm{N}\)
- \(g = 9.81\,\mathrm{m/s^2}\)
Step 1. Resolve the force
\[
\begin{aligned}
F_x &= 20\cos(30^\circ) \approx 17.32\,\mathrm{N} \\
F_y &= 20\sin(30^\circ) = 10.00\,\mathrm{N}
\end{aligned}
\]
Step 2. Compute weight and normal force
\[
\begin{aligned}
W &= mg = (5)(9.81)=49.05\,\mathrm{N} \\
N &= mg - F_y = 49.05 - 10.00 = 39.05\,\mathrm{N}
\end{aligned}
\]
Step 3. Compute friction
\[
f_k = \mu_kN = (0.20)(39.05)=7.81\,\mathrm{N}
\]
Step 4. Add horizontal forces
\[
\begin{aligned}
\sum F_x &= F_x - f_k - D \\
&= 17.32 - 7.81 - 1.50 \\
&= 8.01\,\mathrm{N}
\end{aligned}
\]
Step 5. Compute acceleration
\[
a_x = \frac{8.01}{5} \approx 1.60\,\mathrm{m/s^2}
\]
10. Common mistakes
-
Using \(mg\) instead of \(N\) in friction:
friction is \(f_k=\mu_kN\), not always \(\mu_kmg\).
-
Ignoring the angle:
angled forces must be resolved into components.
-
Using the wrong sign for \(F_y\):
upward pull reduces \(N\), while downward push increases \(N\).
-
Forgetting resistance direction:
friction and drag oppose the direction of sliding.
-
Allowing negative normal force as a contact force:
if the raw normal force is negative, the block has lost contact.
11. Summary
To solve a horizontal sliding problem, resolve the applied force, compute the normal force, find kinetic friction, add horizontal forces, and divide by mass. The calculator follows these steps and connects the algebra to a free-body diagram so the effect of angle, friction, drag, and contact force is visible.
