The diagram is that of a light inextensible string of length 4.2m, whose ends are attached to two fixed points X and Y, 3m apart, and on the same horizontal...

Answer Details

(a) To find the angle between the string and the horizontal, we can use the law of cosines:

c^2 = a^2 + b^2 - 2ab cos C

where c is the length of the string (4.2m), a is the distance from Y to the body (2.4m), and b is the distance from X to Y (3m).

We can rearrange the equation to solve for cos C:

cos C = (a^2 + b^2 - c^2) / 2ab

cos C = (2.4^2 + 3^2 - 4.2^2) / 2 * 2.4 * 3 = 0.8

Finally, we can find the angle using the inverse cosine function:

C = arccos 0.8 = 37.38°

So the angle between the string and the horizontal is 37.38°.

(b) To find the magnitude of the force P, we can use Newton's second law (F = ma), where F is the net force, m is the mass of the body (800g = 0.8kg), and a is the acceleration due to gravity (9.8m/s^2).

The net force is equal to the mass of the body multiplied by the acceleration due to gravity, so:

F = ma = 0.8 * 9.8 = 7.84N

So the magnitude of the force P is 7.84N.

(c) To find the tension T in the string, we can use the fact that the tension must be equal in both parts of the string to keep the system in equilibrium.

Let's call the tension in the part of the string from Y to O "T1" and the tension in the part of the string from O to X "T2".

T1 * cos 37.38° = T2 * cos (180° - 37.38°) = T2 * cos 142.62°

T1 = T2 * cos 142.62° / cos 37.38°

Since T1 = T2, we can substitute T2 for T1:

T2 = T2 * cos 142.62° / cos 37.38°

So, T2 / T2 = cos 142.62° / cos 37.38°

Since T2 is not equal to 0, we can divide both sides by T2:

1 = cos 142.62° / cos 37.38°

To find the tension T in the string, we can use either T1 or T2, since they are equal. Let's use T1:

T1 = T2 = 7.84 / (cos 142.62° / cos 37.38°) = 7.84 / 0.19 = 41.16 N

So the tension in the string is 41.16 N.