An object of mass 100 kg is suspended by two identical cables. The tension in cable one is (400i + 300j)N, where i and j are horizontal and vertical unit vectors respectively. The tension in cable two is (pi + qj)N.
If the object remain at rest, find p and q
To find the values of p and q, we need to consider the equilibrium of forces acting on the object. Since the object is at rest, the sum of the forces in both the horizontal and vertical directions must be zero.
Let's break down the given tensions into their horizontal and vertical components:
Tension in cable one: T1 = (400i + 300j) N
Tension in cable two: T2 = (pi + qj) N
Now, let's find the horizontal and vertical components of each tension:
Horizontal component of T1: T1x = 400 N
Vertical component of T1: T1y = 300 N
Horizontal component of T2: T2x = p N
Vertical component of T2: T2y = q N
Now, we can set up the equilibrium equations:
Sum of horizontal forces (Fx):
T1x + T2x = 0
400 N + p N = 0
This equation tells us that the sum of the horizontal components of the tensions must be zero, indicating that the forces are balanced in the horizontal direction.
Sum of vertical forces (Fy):
T1y + T2y = 0
300 N + q N = 0
This equation tells us that the sum of the vertical components of the tensions must be zero, indicating that the forces are balanced in the vertical direction.
Solving the equations:
From the first equation, p = -400 N.
From the second equation, q = -300 N.
Therefore, p = -400 N and q = -300 N.