A 130 N block rests on a table. The suspended
mass has a weight of 63 N.
130 N
63 N
µs
a) What is the magnitude of the minimum
force of static friction required to hold both
blocks at rest?
Answer in units of N
m1=130/9.8 kg , m2=63/9.8 =
m1•a=0=T-F(fr)
m2•a=0= m2g-T => T=m2•g
F(fr) =T =m2•g
To determine the magnitude of the minimum force of static friction required to hold both blocks at rest, we need to consider the weight of the suspended mass and the coefficient of static friction.
The weight of the suspended mass is given as 63 N. This means that there is a downward force of 63 N acting on the block.
Now, let's calculate the force of static friction. The force of static friction can be determined using the equation:
Force of static friction (Fs) = coefficient of static friction (µs) * normal force (N)
The normal force experienced by the block resting on the table is equal to its weight, which is 130 N.
Now, we can substitute the given values into the equation to calculate the force of static friction:
Fs = µs * N
Fs = µs * 130 N
Since we are looking for the magnitude of the minimum force of static friction, we need to find the maximum value of the coefficient of static friction. The maximum value of the coefficient of static friction is often denoted as µs-max.
However, since the coefficient of static friction is not given in the question, we cannot determine the exact value of the magnitude of the minimum force of static friction without knowing the specific value of the coefficient of static friction.