Question

1. Write a realistic problem for which these are the correct equations.
2. Draw the free-body diagram and the pictorial representation for your problem.
3. Finish the solution of the problem.

$\left(100\mathrm{N}\right)\cos {30}^{\circ }-f_{\mathrm{k}}=\left(20\mathrm{kg}\right)a_x$

$\begin{array}{rl}& n+\left(100\mathrm{N}\right)\sin {30}^{\circ }-\left(20\mathrm{kg}\right)(9.80\mathrm{m}/{\mathrm{s}}^2)=0\\ & f_{\mathrm{k}}=0.20n.\end{array}$

Short answer

Part (a): Suppose we are pulling a block of mass $20\mathrm{kg}$on a rough surface. The coefficient of kinetic friction between the surface is $\mu =0.20$. The force applied on the block is $100\mathrm{N}$at an angle ${30}^{\circ }$above the horizontal. What is the frictional force on the block?

Part (b): The free-body diagram is:

Part(c): The values of $n,f_k$and $a_x$are calculated by solving the given equations. The results are $n=146N$, $f_k=29.2\mathrm{N}$and$a_x=2.87\mathrm{m}/{\mathrm{s}}^2$.

Step-by-step solution

Part (a) Step 1: Given.

Three given equations are:

$\left(100\mathrm{N}\right)\cos {30}^{\circ }-f_k=\left(20\mathrm{kg}\right)a_x$

$n+\left(100\mathrm{N}\right)\sin {30}^{\circ }-\left(20\mathrm{kg}\right)(9.80\mathrm{m}/{\mathrm{s}}^2)=0$

$f_k=0.20n$.

Part (a) Step 2: Explanation of solution.

The real problem is:

Suppose we are pulling a block of$20\mathrm{kg}$mass on a rough surface. The coefficient of kinetic friction between the surface is $\mu =0.20$. The force applied on the block is $100\mathrm{N}$at an angle ${30}^{\circ }$above the horizontal. What is the frictional force on the block?

The three variables in the given equations $n,a_x,f_k$can be represented by the normal force, acceleration of the block, and frictional force. Hence, the three equations can represent the above problem.

Part (b): Step 3: Explanation of solution.

The first equation can represent the total force applied to the block. Which is equal to the difference between $t$frictional force $f_k$and the horizontal component of the applied force.

Since, the block is moving only in $x$direction, the vertical forces are balanced, which is represented by the second equation.

Third equation gives the frictional force $f_k$.

The free-body diagram is:

Part (c): Step 4: Given.

$\begin{array}{rl}& \left(100N\right)\cos {30}^{\circ }-f_k=\left(20\mathrm{kg}\right)a_x\\ & n+\left(100N\right)\sin {30}^{\circ }-\left(20\mathrm{kg}\right)(9.80\mathrm{m}/{\mathrm{s}}^2)=0\\ & f_k=0.20n\end{array}$

Part (c): Step 5: Calculation.

First solve the second equation to calculate $n$.

role="math" localid="1647867990625" $\begin{array}{c}n=\left(20\mathrm{kg}\right)(9.8\mathrm{m}/{\mathrm{s}}^2)-\left(100\mathrm{N}\right)\sin {30}^{\circ }\\ =146\mathrm{N}.\end{array}$

Now, substitute the value of $n$in the third equation to calculate the value of $f_k$.

$\begin{array}{rl}{f}_k=& 0.20\left(146\mathrm{N}\right)\\ & =29.2\mathrm{N}.\end{array}$

Now, use the value of$f_k$in the first equation to calculate the value of $a_x$.

$\begin{array}{rl}{a}_x=& \frac{\left(100\mathrm{N}\right)\cos {30}^{\circ }-\left(29.2\mathrm{N}\right)}{\left(20\mathrm{kg}\right)}\\ & =2.87\mathrm{m}/{\mathrm{s}}^2.\end{array}$

Conclusion,

Part (a): Suppose we are pulling a block of mass $20\mathrm{kg}$on a rough surface. The coefficient of kinetic friction between the surface is $\mu =0.20$. The force applied on the block is $100\mathrm{N}$at an localid="1647868478731" ${30}^{\circ }$angle above the horizontal. What is the frictional force on the block?

Part (b): The free body diagram is:

Part (c): The values of $n,f_k$and $a_x$are calculated by solving the given equations. The results are$n=146N$, $f_k=29.2\mathrm{N}$and $a_x=2.87\mathrm{m}/{\mathrm{s}}^2$.