SME
Statics · Problem type
A frame: the forces in a pair of pliers
Use this for “You squeeze pliers with 50 N, 120 mm from the pin, and the jaws grip 30 mm from it. Find the jaw force and the pin force.”
Diagram
ΣM
=
0
about
the
pin:
F₁d₁
=
F₂d₂,
so
F₁
=
F₂
×
d₂
d₁
=
50
×
120
30
=
200
N
ΣF = 0 on the half: Fₚ = F₁ + F₂ = 200 + 50 = 250 N
The
jaws
grip
d₂
d₁
=
4
times
as
hard
as
the
hand
squeezes.
This is an example. Type your own values to start a new problem.
F₁ (N)
=
F₂ (N)
×
d₂ (mm)
d₁ (mm)
Fₚ (N)
=
F₁ (N)
+
F₂ (N)
Formulas
F₁
=
F₂
×
d₂
d₁
200
=
50
×
120
30
Fₚ
=
F₁
+
F₂
250 = 200 + 50
Assumptions
•
Take one half of the pliers apart: it is a lever on the pin, so ΣM = 0 about the pin.
•
The grip F₂ and the jaw force F₁ are square to the handle, on opposite sides of the pin.
•
Both push the half the same way, so ΣF = 0 makes the pin push back with their sum.
Step-by-step
WE KNOW
F₂ = 50 N, d₂ = 120 mm, d₁ = 30 mm
FIND
jaw force (F₁), pin force (Fₚ)
Step 1 · Find jaw force (F₁)
Use F₁ = F₂ × d₂ ÷ d₁
ΣM
=
0
about
the
pin:
F₁
d₁
=
F₂
d₂
,
so
divide
by
the
jaw
arm.
F₁
=
50
×
120
30
F₁
=
6,000
30
F₁
=
200
N
Step 2 · Find pin force (Fₚ)
Use Fₚ = F₁ + F₂
ΣF = 0 on the half: the pin pushes back against both forces.
Fₚ = 200 + 50
Fₚ = 250 N
Check
Put the numbers back in. Both sides should match.
200
=
50
×
120
30
✓
250 = 200 + 50 ✓
Topic check
See what you know: 5 questions on this topic
New questions each time. Pass to mark it complete.
Related lessons
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