The body at point 𝐵 is at
equilibrium. Find the magnitude of the force 𝐅
sub one in newtons, rounding your answer to two decimal places.
We begin by recalling that if a
body is in equilibrium, the sum of its forces in the horizontal and vertical
directions must both equal zero. From the diagram, we can see there
is a force acting vertically downwards of 10 newtons. We have a horizontal force 𝐅 sub
one. We also have a third force 𝐅 sub
two, which is not acting horizontally or vertically. Our first step is therefore to work
out the horizontal and vertical components of this force.
We begin by letting the angle
between the 𝐅 sub two force and the horizontal equal 𝜃. Using our knowledge of right-angle
trigonometry, the vertical component is equal to 𝐅 sub two multiplied by sin 𝜃 and
the horizontal component is equal to 𝐅 sub two multiplied by cos 𝜃. Resolving horizontally, where
forces acting to the right are positive, we have 𝐅 sub two cos 𝜃 minus 𝐅 sub one
equals zero. Adding 𝐅 sub one to both sides of
this equation, we have 𝐅 sub one is equal to 𝐅 sub two multiplied by cos 𝜃. We will call this equation one.
Resolving vertically, where the
positive direction is upwards, we have 𝐅 sub two sin 𝜃 minus 10 is equal to
zero. Adding 10 to both sides of this
equation, we have 𝐅 sub two multiplied by sin 𝜃 is equal to 10. We will call this equation two.
We now have a pair of simultaneous
equations that we can solve by elimination. We can divide equation two by
equation one such that 𝐅 sub two sin 𝜃 over 𝐅 sub two cos 𝜃 is equal to 10 over
𝐅 sub one. On the left-hand side, we can
divide through by 𝐅 sub two. One of our trigonometric identities
states that sin 𝜃 divided by cos 𝜃 is equal to tan 𝜃. This means that our equation
simplifies to tan 𝜃 is equal to 10 divided by 𝐅 sub one. Multiplying both sides of this
equation by 𝐅 sub one and dividing through by tan 𝜃, we get 𝐅 sub one is equal to
10 over tan 𝜃.
At this stage, we don’t appear to
know the value of tan 𝜃. However, returning to our diagram,
we note that we have a right triangle where the side opposite angle 𝜃 is equal to
110 centimeters and the side adjacent to angle 𝜃 has length 40 centimeters. We know that the tangent of any
angle is equal to the opposite over the adjacent. tan 𝜃 is therefore equal to 110
over 40. And dividing both the numerator and
denominator by 10, this is equal to 11 over four. We can therefore calculate 𝐅 sub
one by dividing 10 by 11 over four or 2.75. 𝐅 sub one is equal to 40 over 11
or 3.6363 and so on. We are asked to round our answer to
two decimal places. The magnitude of the force 𝐅 sub
one is therefore equal to 3.64 newtons to two decimal places.