Q33P - University Physics with Modern Physics

Question

Two adults and a child want to push a wheeled cart in the direction marked x in Fig. P4.33. The two adults push with horizontal forces ${\overrightarrow{\mathbf{F}}}_{\mathbf{1}}$ and ${\overrightarrow{\mathbf{F}}}_{\mathbf{2}}$ as shown. (a) Find the magnitude and direction of the smallest force that the child should exert. Ignore the effects of friction. (b) If the child exerts the minimum force found in part (a), the cart accelerates at $\mathbf{2}\mathbf{.}\mathbf{0}\mathbf{m}\mathbf{/}{\mathbf{s}}^{\mathbf{2}}$ in the +x direction. What is the weight of the cart?

Step-by-Step Solution

Answer

(a) The magnitude and direction of the smallest force that the child should exerts are 16.6 N and at $90°$from the +x axis in the clockwise direction respectively.

 

(b) The weight of the cart is 839.9 N.

Step 1: Identification of the given data:

The given data can be listed below as,

The acceleration of the cart is $\mathrm{a}=2.0\hspace{0.33em}\mathrm{m}/{\mathrm{s}}^2$.

The magnitude of first force is ${\mathrm{F}}_1=100 \mathrm{N}$.

The magnitude of the second force is ${\mathrm{F}}_2=140 \mathrm{N}$.

The angle of inclination of the first force from the +x axis is  ${\theta }_1=60°$.

The angle of inclination of the second force from the +x axis is ${\theta }_2=30°$.

Step 2: Significance of resultant force

Whenever some external force acts on an object at different angles of inclination, the effects of all forces along the horizontal and vertical direction refer to the resultant force along the horizontal and vertical direction.

Step 3: (a) Determination of the magnitude of the smallest force that the child should exerts:

The relation of magnitude of the smallest force that the child should exerts is expressed as,

 

$\mathrm{F}=\left[\left({\mathrm{F}}_1{\mathrm{sin\theta }}_1\right)-\left({\mathrm{F}}_2{\mathrm{sin\theta }}_2\right)\right]$ 

 

Here, F is the magnitude of the smallest force that the child should exert.

 

Substitute all the known values in the above equation.

$$\begin{gathered} \mathrm{F}=\left[\left(100 \mathrm{N}\right)\left(\sin 60°\right)-\left(140 \mathrm{N}\right)\left(\sin 30°\right)\right] \\ =\left[\left(100\mathrm{N}\right)\left(0.866\right)-\left(140 \mathrm{N}\right)\left(0.5\right)\right] \\ =86.6 \mathrm{N}-70 \mathrm{N} \\ =16.6 \mathrm{N} \end{gathered}$$ 

 

 

The direction of the smallest force that the child should exert would be at $90°$ from the +x axis in the clockwise direction.

 

Thus, the magnitude of the smallest force that the child should exert is 16.6 N and the direction of the smallest force that the child should exert would be at $90°$ from the +x axis in the clockwise direction.

Step 4: (b) Determination of the weight of the cart

The expression to calculate the resultant force in the +x direction is expressed as,

 

${\mathrm{F}}_{\mathrm{x}}=\left[\left({\mathrm{F}}_1{\mathrm{cos\theta }}_1\right)+\left({\mathrm{F}}_2{\mathrm{cos\theta }}_2\right)\right]$ 

 

Here, ${\mathrm{F}}_{\mathrm{x}}$ is the resultant force in the +x direction.

 

Substitute all the known values in the above equation.

$$\begin{gathered} {\mathrm{F}}_{\mathrm{x}}=\left[\left(100\mathrm{N}\right)\left(\cos 60°\right)+\left(140 \mathrm{N}\right)\left(\cos 30°\right)\right] \\ =\left[\left(100\mathrm{N}\right)\left(0.5\right)+\left(140\right)\left(0.866\right)\right] \\ =50 \mathrm{N}+121.24 \mathrm{N} \\ =171.24 \mathrm{N} \end{gathered}$$ 

 

 

The expression to calculate the weight of the cart is expressed as,

 

$$\begin{gathered} \mathrm{W}=\mathrm{mg} \\ =\left(\frac{{\mathrm{F}}_{\mathrm{x}}}{\mathrm{a}}\right)\mathrm{g} \end{gathered}$$ 

 

Here, W is the weight of the cart.

 

Substitute all the known values in the above equation.

$$\begin{gathered} \mathrm{W}=\left(\frac{171.24 \mathrm{N}}{2.0 \mathrm{m}/{\mathrm{s}}^2}\right)\left(9.81 \mathrm{m}/{\mathrm{s}}^2\right) \\ =839.9 \mathrm{N} \end{gathered}$$ 

 

 

Thus, the weight of the cart is 839.9 N.