A 32 kg child puts a 15 kg box into a 12 kg wagon. The child then pulls horizontally on the wagon with a force of 65 N. if the box does not move relative to the wagon, what is the static friction force on the box?
Fap - Fs = 0,
65 - Fs = 0
Fs = 65 N. = Force of static friction.
fx=22cos35...18n
MY BAD WRONG QUESTION
To find the static friction force on the box, we need to first understand the concept of static friction.
Static friction is the force that opposes the motion between two surfaces that are in contact with each other but are not moving relative to each other. It keeps the object at rest until an external force is applied to overcome it.
In this scenario, the box is not moving relative to the wagon. This implies that the force applied by the child is balanced by the static friction force applied by the wagon on the box.
To calculate the static friction force, we can use the equation:
Fs = μs * N
where Fs is the static friction force, μs is the coefficient of static friction, and N is the normal force.
In this case, the normal force (N) acting on the box is equal to the weight of the wagon plus the weight of the box. The weight is calculated by multiplying the mass (m) by the acceleration due to gravity (g).
N = (m_wagon + m_box) * g
where m_wagon is the mass of the wagon, m_box is the mass of the box, and g is the acceleration due to gravity (approximately 9.8 m/s^2).
N = (12 kg + 15 kg) * 9.8 m/s^2
= 27 kg * 9.8 m/s^2
≈ 264.6 N
Now, to find the coefficient of static friction (μs), we can rearrange the equation as:
Fs = μs * N
μs = Fs / N
Since the static friction force (Fs) is equal to the force applied by the child of 65 N, we can substitute the values into the equation:
μs = 65 N / 264.6 N
≈ 0.245
Therefore, the static friction force on the box is approximately 0.245 times the normal force:
Fs = 0.245 * 264.6 N
≈ 64.8 N
So, the static friction force on the box is approximately 64.8 N.