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Engineering Statics: Open and Interactive

Section 9.3 Wedges

A wedge is a tapered object which converts a small input force into a large output force using the principle of an inclined plane. Wedges are used to separate, split or cut objects, lift weights, or fix objects in place. The mechanical advantage of a wedge is determined by the angle of its taper; narrow tapers have a larger mechanical advantage.
Wedges are used in two primary ways:
Low friction wedges are a simple machines which allows users to create large output forces to move objects using comparatively small input forces. In the log splitter in Figure 9.3.1.(a), hydraulic ram pushes a log into a stationary wedge. The normal force pushes the two halves of the log apart while the friction force \(F_f\) is opposes the pushing force \(P\text{.}\)
High-friction (self-locking) wedges control the location of objects or hold them in place. Examples include doorstop wedges and carpentry wedges. The sailor in Figure 9.3.1.(b) is hammering two wooden wedges towards each other to create large compressive forces to secure shoring timbers during a damage control operation.
Image of a hydraulic ram splitting a piece of wood.
(a) A low friction wedge is used to split logs.
Sailors pounding wedges into damage control shoring.
(b) High friction wedges are used to secure shoring timbers.
Figure 9.3.1. Wedges in use.
Luckily the analysis of low- and high-friction wedges are identical and they are quite similar to the multi-force body equilibrium problems we saw in Chapter 5 and Chapter 6. The main difference is the inclusion of friction from all non-smooth contact surfaces. The directions of both the normal and friction forces on the free-body diagrams are defined below.
Normal forces act between bodies act perpendicular to the contacting surfaces. All normal forces on a free-body diagram should be pointing towards the body because wedges are always subjected to compression.
Friction forces are between bodies which act parallel or tangential to the contacting surfaces and are created by the microscopic or large scale roughness of the surfaces. All friction forces on a free-body diagram should be drawn pointing in the direction which resists relative motion at the point of contact.
The key added challenge of solving wedge problems is that the angled faces of wedges usually need to be resolved into components in the \(x\) and \(y\text{,}\) unless a different coordinate system is used.
One of the critical steps in solving block or wedge problems is to determine which force is engaging the friction of the system. Start by drawing the friction forces on the body where this force acts. As you pass the friction and normal forces to adjacent free-body diagrams, you must always show them as equal and opposite, action-reaction pairs. This is illustrated in the following example.

Example 9.3.2.

Find the minimum force \(P\) required to start raising the \(\lb{10}\) block using a 25° wedge.
Assume that the coefficient of static friction at both contact surfaces is \(\mu_s = 0.20\) and that the wedge is massless.
Schematic of a force being applied to an angled wedge to lift a larger block against a frictionless roller.
\begin{equation*} P = \lb{9.35} \end{equation*}
Given: \(W = \lb{10}, \mu_s = 0.2,\) and \(\theta=\ang{25}\text{.}\)
Start by assuming the wedge is on the verge of moving to the right and drawing free-body diagrams of the system and each of its parts.
(a) Both parts
(b) Block
(c) Wedge
Figure 9.3.3.
By considering equilibrium of the combined wedge and block in the \(y\) direction, we find that
\begin{equation*} N_1 = W = \lb{10} \end{equation*}
Assuming that the wedge is pushed to point of impending motion,
\begin{equation*} F_1 = N_1 \mu_s = \lb{2}\text{.} \end{equation*}
There are still two unused equilibrium equations for this free-body diagram, but they are not sufficient to find unknown forces \(P\) and \(R\) because the point where resultant friction force acts is also unknown.
Next, consider equilibrium of the block, Figure 9.3.3.(b). A detail of \(N_2\) and \(F_2\) in Figure 9.3.4 shows how the their \(x\) and \(y\) components are related to the wedge angle \(\theta\text{.}\)
Applying equilibrium in the \(y\) direction with \(F_2 = \mu_s N_2\) gives
\begin{align*} \Sigma F_y \amp = 0\\ -W + N_{2_y} - F_{2_y} \amp = 0\\ -W + N_2 \cos \theta - \mu_s N_2 \sin \theta \amp = 0\\ N_2(\cos\theta - \mu_s \sin\theta) \amp= W\\ N_2 \amp = \lb{12.17} \end{align*}
Figure 9.3.4.
Since the wedge is at the point of impending motion,
\begin{align*} F_2 \amp= \mu_s N_2 \\ \amp= \lb{2.43} \end{align*}
With the friction and normal forces on both sides of the wedge known, we can now find the required force \(P\) using the free body diagram of the wedge 9.3.3.(c).
\begin{align*} \Sigma F_x \amp = 0\\ P - F_1 - F_{2_x} - N_{2_x} \amp = 0\\ P \amp = F_1 + F_2 \cos \theta + N_2 \sin \theta \\ P \amp = \lb{9.35} \end{align*}

Example 9.3.5.

Using the same system as in Example 9.3.2, find the minimum force \(P'\) to prevent the wedge from slipping out from under the block
This is similar to the previous example, but the free-body diagrams need to change in the following ways:
  • all friction force directions reverse since the impending motions of the block and wedge have reversed, and
  • the direction of \(P'\) may have to reverse if the wedge has sufficient friction to stay static when \(P'=0\text{.}\)
Note that for forces between \(P\) and \(P'\text{,}\) the system is static and the friction forces are static-but-not-impending.
\begin{equation*} P' = \lb{0.94} \end{equation*}
The solution is left as an exercise for the reader.