The system shown in the following figure is in static equilibrium. Find the frictional force exerted on block (m1) given that the mass of block (m1) is 8.90 kg, the mass of block (m2) is 2.59 kg, and the coefficient of static friction between block (m1) and the surface on which it rests is 0.391.
To find the frictional force exerted on block (m1), we can use the equation for static friction:
\(f_s = \mu_s \cdot N\),
where \(f_s\) is the static frictional force, \(\mu_s\) is the coefficient of static friction, and \(N\) is the normal force.
To find the normal force, we need to consider the forces acting on block (m1). In this case, there are two forces acting on the block: the gravitational force (\(F_{\text{gravity}}\)) and the normal force (\(N\)). Since the system is in static equilibrium, the sum of the vertical forces must be zero:
\(F_{\text{gravity}} + N = 0\),
where \(F_{\text{gravity}}\) is given by \(m_1 \cdot g\), where \(m_1\) is the mass of block (m1) and \(g\) is the acceleration due to gravity.
Substituting the value of \(F_{\text{gravity}}\) into the equation, we can solve for the normal force \(N\):
\(m_1 \cdot g + N = 0\),
\(N = -m_1 \cdot g\).
Now that we have the value of the normal force, we can substitute it into the equation for static friction to find the static frictional force \(f_s\):
\(f_s = \mu_s \cdot (-m_1 \cdot g)\).
Substituting the given values:
\(f_s = 0.391 \cdot (-8.90 \, \text{kg} \cdot 9.8 \, \text{m/s}^2)\).
Calculating this expression will give us the value of the frictional force exerted on block (m1).