Question
7. Three forces F_(1)=(2m+n),F_(2)=(-3m+4n) and F_(3)=(4n-6n) act on a body with mass 12 kg. Find the magnitude of the acceleration of the body.
Answer
4.1
(328 Votes)
Yves
Advanced · Tutor for 1 years
Answer
The magnitude of acceleration could be found as: a = F_net/m = (3i -1j)/12
Explanation
To find acceleration, we'll need Newton's second law "F = m * a" which states that the force applied on an object is equal to mass times acceleration. Here in the given situation, three forces are acting on the body and according to law of vector addition, the net force acting on body is : F_net = F1 + F2 + F3. Characteristic specifications of i and j directions are mapped respectively-i=2m and j=n1) F1 = (2m+n) = 2i +j2) F2 = (-3m+4n) = -3i + 4j3) F3 = (4m-6n) = 4i -6jF_net = (F1 + F2 + F3) = (2 - 3 + 4)i + (1 + 4 - 6)j = 3i -1jThe mass of the object (m) is given as 12 kg. We can use the link between force and acceleration (F = m * a) to establish the magnitude of the acceleration vector by setting the acceleration (a) constituent to net force (F_net) divided by the mass.