The following questions appeared in KEAM 2007 (Engineering) question paper. Questions of this type are often found in various Medical and Engineering Entrance examinations and other similar examinations such as AP Physics Exam and GRE.

**(1) A mass of 6kg is suspended by a rope of length 2m from a ceiling. A force of 50 N in the horizontal direction is applied at the mid point of the rope. The angle made by the rope with the vertical, in equilibrium is **

**(a) 50º** **(b) 60º** ** (c) 30º** **(d) 40º** **(e) 45º** ** **

This can be easily worked out using tangent law in Statics dealing with concurrent forces. With reference to the figure, we have

tanθ = F/W

where F is the horizontal force, W is the *weight* (in newton) suspended and θ is the angle made by the rope with the vertical. [ Note that the tangent law follows from the law of triangle of forces].

In the present case, F = 50 N and W= mg.

Even if you take the value of ‘g’ to be approximately equal to 10 ms^{–2}, you will arrive at the answer. Thus W = 6×10 = 60 N so that tanθ = 50/60 = 5/6.

Even though you are not allowed to use physical and mathematical tables at the examination, you can arrive at the correct option as 40º since you know that tan 30º = 1/√3 and tan 45º = 1.

**(2) The sum of the magnitudes of two forces acting at a point is 18 N and the magnitude of their resultant is 12 N. If the resultant is at 90º with the smaller force, the magnitude of the forces in newton are**

**(a) 6, 12 (b) 11, 7 (c) 5, 13 (d) 14, 4 (e) 10, 8**

If P and Q are the two forces, we have P + Q = 18 N so that P = 18 – Q.

Since the resultant R (12 N) is at right angles to the smaller force P (fig.) we can write

Q = √(P^{2} + 12^{2}) = √[(18–Q)^{2} + 12^{2}], from which Q = 13 N.

Therefore, P = 18–13 = 5 N.

The correct option therefore is (c).

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