Question

Three concurrent co-planar forces $1\,N, 2\,N$ and $3\,N$ acting along different directions on a body

• A. can keep the body in equilibrium if 1 N and 2 N act at right angles.
• B. cannot keep the body in equilibrium
• C. can keep the body in equilibrium if 1 N and 3 N act at an acute angle.
• D. can keep the body in equilibrium if 2 N and 3 N act at right angles

Answer 1

The Correct Option is B

If we keep $1\, N$ and $2\, N$ forces act in same direction then these are balanced by $3\, N$ force, but this is against statement of question.

Answer 2

Let,

\(\vec{F_1}\)=1N

\(\vec{F_2}\)=2N

\(\vec{F_3}\)=3N

We have to find in which condition we can keep the body in equilibrium.

Body in equilibrium when the sum of all the forces acting on a body is zero

∑F→=0

\(\vec{F_1}\)+\(\vec{F_2}\)+\(\vec{F_3}\)=0

It is only possible when

∣\(\vec{F_1}\)+\(\vec{F_2}\)∣ = ∣\(\vec{F_3}\)∣

Here,

\(\vec{F_1}\)=1N

\(\vec{F_2}\)=2N

\(\vec{F_3}\)=3N

Hence, \(\vec{F_1}\)and \(\vec{F_2}\)

should act in one direction and \(\vec{F_3}\)

should act in the opposite direction of \(\vec{F_1}\)and \(\vec{F_2}\)

But the problem statement says that three concurrent coplanar forces \(\vec{F_1}\) , \(\vec{F_1}\) , and \(\vec{F_3}\)

acting along different directions on a body. But when \(\vec{F_1}\)and \(\vec{F_2}\)

act along the same direction then only the body will be in equilibrium. If \(\vec{F_1}\)and\(\vec{F_2}\)

act along different directions then we cannot keep the body in equilibrium.

Hence, option (B) is correct.

Answer 3

How can a body remain in equilibrium under the action of three contemporaneous coplanar forces of magnitudes 1N, 2N, and 3N operating in opposing directions? There are two methods for answering these kinds of queries. One is to get the solution right away. Another alternative is to examine each of the available choices individually to see which one produces the desired outcome. Let's use the second approach as it appears to be the simpler one.

Verify the first selection. Given that 2N and 3N will operate at right angles, the body will be in equilibrium. Assume that the 1N force operates at any arbitrary location whereas the 2N, 3N, and 5N forces act at a straight angle. The force that results will be 

\(\sqrt{2^2+3^2+2(2)(3)cos(90)}\)

=\(\sqrt{4+9+0}\)

=\(\sqrt{13}\) N

As a result, there are now two forces: 1N and 13N. Two forces can only be in balance with one another if their respective magnitudes and directions are equal. Thus, the body cannot be in homeostasis under this circumstance.

Let's examine the second choice. similar steps as those for option one should be taken. The resulting force in this case will be \(\sqrt{2^2+3^2+2(2)(3)cos(90)}\)=1+4+0=5.

3N and \(\sqrt{5N}.3\) are the only two forces left at this point and 3 \(\neq\) \(\sqrt{5}\) , therefore the body won't be in balance.

Let's save the choice B for later. Option D states that if the forces of 1N and 3N operate at an acute angle, the body can be in equilibrium. Let's say that 1N and 3N act at an angle.

 

These two add up to \(\sqrt{1^2+3^2+2(1)(3)cos \theta}\) = \(\sqrt{10+6cos\theta}\) 

\(\Rightarrow\) 4=10+6cos. Solve for θ

\(\Rightarrow\)−6=6cosθ

\(\Rightarrow\)cosθ=−1.

Hence \(\theta\) =\(\pi\) or 180°. The angle is not sharp, though. This makes the second choice inaccurate as well. The right response is (B), which cannot maintain the body's equilibrium.