Select the correct answer.
Mary pushes a crate by applying force of 18 newtons. Unable to push it alone, she gets help from her friend, Anne. Together they apply a force of 43 newtons, and the crate just starts moving. If the coefficient of static friction is 0.11, what is the value of the normal force?
A. [tex]$2.0 \times 10^2$[/tex] newtons
B. [tex]$2.6 \times 10^2$[/tex] newtons
C. [tex]$3.9 \times 10^2$[/tex] newtons
D. [tex]$4.0 \times 10^2$[/tex] newtons
E. [tex]$4.3 \times 10^2$[/tex] newtons
Answer (1)
The problem involves finding the normal force given the applied force and the coefficient of static friction.
The force of static friction equals the applied force when the crate starts moving: F f = 43 N .
The coefficient of static friction is given as μ s = 0.11 .
Calculate the normal force using the formula N = μ s F f = 0.11 43 ≈ 390.91 N .
The closest answer choice is 3.9 × 1 0 2 newtons .
Explanation
Understanding the Problem We are given that Mary and Anne together apply a force of 43 newtons to a crate, and this is just enough to start the crate moving. This means that the applied force is equal to the maximum force of static friction. We are also given that the coefficient of static friction is 0.11. Our goal is to find the normal force.
Relating Friction and Normal Force The force of static friction ( F f ) is related to the normal force ( N ) and the coefficient of static friction ( μ s ) by the equation: F f = μ s N
Identifying Given Values Since the applied force is equal to the force of static friction when the crate just starts moving, we have: F a ppl i e d = F f = 43 N μ s = 0.11
Calculating Normal Force We can rearrange the equation F f = μ s N to solve for the normal force N :
N = μ s F f = 0.11 43 ≈ 390.91 N
Selecting the Correct Answer The normal force is approximately 390.91 N. Looking at the answer choices, the closest value is 3.9 × 1 0 2 newtons.
Examples
Understanding static friction and normal force is crucial in many real-world scenarios, such as designing brakes for vehicles. The brakes need to apply enough force to stop the vehicle from moving, and this force depends on the coefficient of friction between the brake pads and the wheels, as well as the normal force. By calculating the required normal force, engineers can design effective braking systems that ensure safety. For example, if a car weighs 1500 kg and the coefficient of static friction between the tires and the road is 0.8, the maximum static friction force is calculated as follows: Weight = 1500 kg × 9.8 m/s 2 = 14700 N . Normal force (N) is equal to the weight on a flat surface, so N = 14700 N . The maximum static friction force is F f = μ s N = 0.8 × 14700 N = 11760 N . This calculation helps determine the braking force needed to prevent skidding.
