Q26P Figure 6-32 shows three crates b... [FREE SOLUTION]

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Chapter 6: Q26P (page 142)

Figure 6-32 shows three crates being pushed over a concrete floor by a horizontal force of magnitude $440 N$ . The masses of the crates are $m_{1} = 30.3 kg$ , $m_{2} = 10.1 kg$ , and $m_{2} = 20.0 kg$ .The coefficient of kinetic friction between the floor and each of the crates is $0.700$ . (a) What is the magnitude $F_{32}$ of the force on crate 3 from crate 2? (b) If the crates then slide onto a polished floor, where the coefficient of kinetic friction is less than $0.700$ , is magnitude $F_{32}$ more than, less than, or the same as it was when the coefficient was $0.700$ ?

Short Answer

Expert verified

(a) The magnitude $F_{32}$ of the force on crate 3 from crate 2 is $147 N$ .

(b) The magnitude of $F_{32}$ remains the same.

Step by step solution

Given

$m_{1} = 30.0 kg m_{2} = 10.0 kg m_{3} = 20.0 kg$

Force, $F = 440 N$

Coefficient of kinetic friction between the Crates and floor, $渭_{k} = 0.700$

Determining the concept

To find the force on box 3 due to box 2, find the acceleration of the system and then by using the Newton鈥檚 2nd law of motion, the force $^{F_{32}}$ can be found. According to Newton's 2nd law of motion, a force applied to an object at rest causes it to accelerate in the direction of the force.

Formula:

$F_{n e t} = 鈭� m a$

where, F is the net force, m is mass and a is an acceleration.

Determining the free body diagram

Free Body Diagram of the system $(m_{1} + m_{2} + m_{3})$ and 3rd crate:

(a) Determining the magnitude F32of the force on crate 3 from crate 2

By using Newton鈥檚 2nd law of motion along the vertical direction to the system,

$N - (m_{1} + m_{2} + m_{3}) g = 0 N = (m_{1} + m_{2} + m_{3}) g$

Thus, the frictional force,

$f_{k} = 渭_{k} N = 渭_{k} (m_{1} + m_{2} + m_{3}) g$

Now, applying Newton鈥檚 2nd along horizontal direction to the system,

${f_{k}}^{'} = 渭_{k} N^{'} = 渭_{k} m_{3} g F_{32} - {f_{k}}^{'} = m_{3} a F_{32} = m_{3} a + {f_{k}}^{'} = (20.0) (0.47) + (0.700) (20.0) (9.81) = 147 N$

Using the Newton鈥檚 2nd law along vertical and horizontal direction to the 3rd crate,

$N' - m_{3} g = 0 N' = m_{3} g$

Then frictional force,

${f_{k}}^{'} = 渭_{k} N^{'} = 渭_{k} m_{3} g F_{32} - {f_{k}}^{'} = m_{3} a F_{32} = m_{3} a + {f_{k}}^{'} = (20.0) (0.47) + (0.700) (20.0) (9.81) = 147 N$

Hence, the magnitude $^{F_{32}}$ of the force on crate 3 from crate 2 is $^{147 N}$

(b) Determining if the magnitude F32is more than, less than, or the same as it was when the coefficient was 0.700

As from the above,

$F_{32} = m_{3} a + {f_{k}}^{'} = m_{3} ((\frac{F - 渭_{k} (m_{1} + m_{2} + m_{3}) g}{(m_{1} + m_{2} + m_{3})}) + 渭_{k} g) = m_{3} (\frac{F - 渭_{k} (m_{1} + m_{2} + m_{3}) g + 渭_{k} (m_{1} + m_{2} + m_{3}) g}{(m_{1} + m_{2} + m_{3})}) = m_{3} (\frac{F}{(m_{1} + m_{2} + m_{3}})$