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Chapter 4: Problem 83

Two blocks are stacked on a frictionless table, and a horizontal force \(F\) is applied to the top block (block 1). Their masses are \(m_{1}=2.50 \mathrm{~kg}\) and \(m_{2}=3.75 \mathrm{~kg}\). The coefficients of static and kinetic friction between the blocks are 0.456 and 0.380 , respectively a) What is the maximum applied force \(F\) for which \(m_{1}\) will not slide off \(m_{2} ?\) b) What are the accelerations of \(m_{1}\) and \(m_{2}\) when \(F=24.5 \mathrm{~N}\) is applied to \(m_{1}\) ?

Short Answer

Expert verified
To find the maximum force, use the formula: $$ F_{max} = f_{s,max} = μ_s m_{1}g. $$ Plug in the values for \(μ_s\), \(m_{1}\), and \(g\). b) What are the accelerations of the two blocks when a force of 24.5 N is applied to the top block? To find the accelerations of the two blocks, first find the kinetic friction force: $$ f_k = μ_k m_{1}g. $$ Then, find the net forces acting on each block: $$ F_{net,1} = F - f_k \quad \text{and} \quad F_{net,2} = f_k. $$ Finally, use Newton's second law to find the accelerations: $$ a_{1} = \frac{F_{net,1}}{m_{1}} \quad \text{and} \quad a_{2} = \frac{F_{net,2}}{m_{2}}. $$ Calculate the values for \(a_{1}\) and \(a_{2}\).

Step by step solution

01

1. Calculate the maximum static friction force between the blocks

The maximum static friction force \(f_{s,max}\) between the two blocks is given by the product of the coefficient of static friction \(μ_s\) and the normal force \(N\). Since there is no vertical acceleration, the normal force between the blocks must equal the weight of block 1: \(N = m_{1}g\). Thus, $$ f_{s,max} = μ_s m_{1}g, $$ where \(g = 9.81 \,\text m/s^2\) is the gravitational acceleration.
02

2. Find the maximum force F

For block 1 to not slide off block 2, the applied force should not exceed the maximum friction force between the blocks. Therefore, $$ F_{max} = f_{s,max} = μ_s m_{1}g. $$ Plug in the values for \(μ_s\), \(m_{1}\), and \(g\) to find the maximum applied force \(F_{max}\).
03

3. Calculate the kinetic friction force between the blocks

When a force of \(F=24.5\,\text N\) is applied to block 1, the kinetic friction force \(f_k\) between the blocks needs to be calculated. Since the force is less than the maximum force calculated in step 2, the blocks will move together. The kinetic friction force can be calculated by the product of the coefficient of kinetic friction \(μ_k\) and the normal force: $$ f_k = μ_k N = μ_k m_{1}g. $$
04

4. Calculate the total force acting on each block

The total net force acting on block 1, \(F_{net,1}\), will be the applied force minus the kinetic friction force: $$ F_{net,1} = F - f_k. $$ The total net force acting on block 2, \(F_{net,2}\), will be equal to the kinetic friction force: $$ F_{net,2} = f_k. $$
05

5. Calculate the accelerations of the two blocks

Using Newton's second law, the acceleration of each block can be found as follows: $$ a_{1} = \frac{F_{net,1}}{m_{1}} \quad \text{and} \quad a_{2} = \frac{F_{net,2}}{m_{2}}. $$ Plug in the values for \(F_{net,1}\), \(F_{net,2}\), \(m_{1}\), and \(m_{2}\) to find the accelerations \(a_{1}\) and \(a_{2}\) of the blocks.

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