Two crates, of mass 65 kg and 125 kg, are in contact and at rest on a horizontal surface. A 650-N force is exerted on the 65-kg crate. If the coefficient of kinetic friction is 0.18, calculate the force that each crate exerts on the other.
typo 428 N
Determine do they accelerate ?
weight = m g = (65+125)(9.8) = 1862 N
so kinetic friction = .18*1862 = 335 Newtons
so
a = (650-335)/(190) = 1.66 N/m^2
Force on big crate from little crate - 0.18*125*9.8 = 125*1.66
so F on big = 207.5+ 220.5 = 428 N
OR
just say F = [125/(125+65) ]* 650 = 438 N
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To calculate the force that each crate exerts on the other, we need to use the concept of Newton's third law - for every action, there is an equal and opposite reaction.
First, let's calculate the force of friction acting on the 65-kg crate. The formula for the force of friction is given by:
\( F_{friction} = \mu_k \cdot F_N \)
Where \( F_N \) is the normal force exerted on the crate and \( \mu_k \) is the coefficient of kinetic friction.
The normal force \( F_N \) can be calculated as the weight of the crate, which is given by:
\( F_N = m \cdot g \)
where \( m \) is the mass of the crate and \( g \) is the acceleration due to gravity.
In this case, the mass of the 65-kg crate is given, and the acceleration due to gravity is approximately 9.8 m/s^2. Thus, we can calculate \( F_N \) as:
\( F_N = 65 \, \text{kg} \cdot 9.8 \, \text{m/s}^2 \)
Next, we can plug the value of \( F_N \) and the coefficient of kinetic friction (\( \mu_k \)) into the formula to calculate the force of friction.
\( F_{friction} = 0.18 \cdot F_N \)
Now that we have the force of friction acting on the 65-kg crate, we can use Newton's third law to determine the force exerted by the 65-kg crate on the 125-kg crate. According to the third law, this force will be equal in magnitude and opposite in direction to the force of friction.
So, the force exerted by the 65-kg crate on the 125-kg crate is \( F_{friction} = 0.18 \cdot F_N \).
Finally, to determine the force exerted by the 125-kg crate on the 65-kg crate, we can use Newton's third law, which tells us that this force will be equal in magnitude and opposite in direction to the force exerted by the 65-kg crate on the 125-kg crate.
Thus, the force exerted by the 125-kg crate on the 65-kg crate is also \( F_{friction} = 0.18 \cdot F_N \).