Question

a 1 kg box is pushed against a wall horizontally with the minimum force required to keep the box at rest, 12.6 newtons. determine the coefficient of static friction for the box.question 81 ptsa yachtsman wants to modify his cooler so that it wont slide around the deck of his boat when it turns suddenly. which of the following things could the yachtsman do that would not change the coefficient of friction between the cooler and the floor?use sandpaper on the bottom of the cooler to make the surface rougherput a heavy weight on top of the coolersprinkle sand underneath the coolerglue a rubberized mat to the bottom of the coolerquestion 91 ptsa 2 kg box is pulled by a tension force of 7.1 newtons at an angle of 40.2 degrees from the horizontal. the box moves across a flat table. the coefficient of kinetic friction between the box and the table is 0.12. calculate the magnitude of the normal force between the box and the table.

Explanation:

Step1: Analyze forces in vertical direction

The box is at rest, so in the vertical direction, the force of static friction \(f_s\) balances the weight \(mg\) of the box. \(f_s = mg\). Also, the formula for static friction is \(f_s=\mu_sN\), where \(N\) is the normal force. Here, the normal force \(N\) is equal to the applied force \(F = 12.6\space N\) (since in the horizontal direction, for equilibrium \(N = F\)). The mass \(m = 1\space kg\) and \(g=9.8\space m/s^2\).

Step2: Calculate the coefficient of static friction

From \(f_s=\mu_sN\) and \(f_s = mg\), we can solve for \(\mu_s\). Substitute \(f_s=mg\) and \(N = F\) into the formula \(\mu_s=\frac{f_s}{N}=\frac{mg}{F}\).
Substitute \(m = 1\space kg\), \(g = 9.8\space m/s^2\) and \(F=12.6\space N\) into the formula: \(\mu_s=\frac{1\times9.8}{12.6}\approx0.78\)

for Question 8:
The coefficient of friction \(\mu\) depends on the nature of the two surfaces in contact. The formula for frictional force is \(f=\mu N\), where \(N\) is the normal force.

• Using sand - paper on the bottom of the cooler, sprinkling sand underneath the cooler and gluing a rubberized mat to the bottom of the cooler all change the nature of the surfaces in contact, thus changing \(\mu\).
• Putting a heavy weight on top of the cooler changes the normal force \(N\) (since \(N=(m_{cooler}+m_{weight})g\)), but does not change the coefficient of friction \(\mu\) (because \(\mu\) is a property of the two surfaces, not of the normal force or the mass).

for Question 9:
Analyze the forces in the vertical direction. The weight of the box is \(W = mg\), where \(m = 2\space kg\) and \(g = 9.8\space m/s^2\), so \(W=2\times9.8 = 19.6\space N\). The vertical component of the tension force is \(F_{y}=F\sin\theta\), where \(F = 7.1\space N\) and \(\theta = 40.2^{\circ}\), so \(F_{y}=7.1\times\sin(40.2^{\circ})\approx7.1\times0.645 = 4.58\space N\).
In the vertical direction, using the equilibrium equation \(N+F_{y}=mg\) (taking upward as positive), we can solve for \(N\).
\(N=mg - F\sin\theta\)
Substitute \(m = 2\space kg\), \(g = 9.8\space m/s^2\), \(F = 7.1\space N\) and \(\theta = 40.2^{\circ}\) into the formula:
\(N=2\times9.8-7.1\times\sin(40.2^{\circ})\)
\(N = 19.6-4.58=15.02\space N\)

Answer:

\(0.78\)