Tension Problem

Question

A block of mass m = 4.50 kg rides on top of a second block of mass M = 15.0 kg. A person attaches a string to the bottom block and pulls the system horizontally across a frictionless surface as in figure (a). Friction between the two blocks keeps the 4.50 kg block from slipping off. If the coefficient of static friction is 0.430, answer the following.

(a) What maximum force can be exerted on the string without causing the 4.50 kg block to slip?

 I got 82.17N

(b) Use the system approach to calculate the acceleration.

 I got 4.21 m/s²

Use the values from above to help you work this exercise. Suppose instead that the string is attached to the top block as shown in figure (b). Find the maximum force that can be exerted by the string on the block without causing the top block to slip.

This is what I need help with.

Answer 1

This question is solvable even without the figure (your referenced figure is missing :-)). I will split the problem into two parts:

1) bottom block, with mass m₁ and 2) top block, with mass m₂

Draw the FBD for both cases and from that, we have the various components of the forces to be

bottom block

---------------------

x: ∑F₁,_x = F₁ - F_s = m₁a_{1,x --------------------------------------1}

y: ∑F_1,y = N₁ - N₂ - m₁g = 0 ==> N₁ = N₂ + m₁g --------------------------------------2

top block

--------------------

x: ∑F₂,_x = F_s = m₂a_{2,x ------------------------------------- 3}

y: ∑F_2,y = N₂ - m₂g = 0 ==> N₂ = m₂g ------------------------------------- 4

b)

Assuming a_2,x = a_1,x = a, then from equation (3), you get

F_s = m₂a = µ_sN₂ = µ_sm₂g ==> a = µ_sg = 0.43(9.8m/s²) = 4.21m/s²

a) From equation (1), we have

F₁ = F_s + m₁a = µ_sg(m₂ + m₁) = 0.43(9.8m/s²)(4.5kg + 15.0kg) = 82.17N

Your results (for the first case) are indeed correct :-)

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Now let's do similar calculation for the case where the applied force is attached to the top block

Split the problem into two parts again: 1) bottom block and 2) top block

bottom block

-------------------------

x: ∑F₁,_x = F_s = m₁a1_{,x ----------------------------------- 11}

_y: ∑F_1,y = N₁ - N₂ -m₁g = 0 ==> N₁ = g(m₂ + m₁) -----------------------------------22

top block

------------------------

x: ∑F₂,_x = F₂ - F_s = m₂a₂,_{x ------------------------------------33}

y: ∑F₂,_y = N₂ - m₂g + 0 ==> N₂ = m₂g ------------------------------------44

b)

Assuming a₂,_x = a_1,x = a, then from equation (11), you get

F_s = m₁a = µ_sm₂g ==> a = µ_sg(m₂/m₁) = 0.43(9.8m/s²)(4.5kg/15.0kg) = 1.26m/s²

a) From equation (33), you have

F₂ = F_s + m₂a = µ_sm₂g + m₂a = µ_sm₂g + m₂µ_sg(m₂/m₁) = m₂µ_sg(1 + m₂/m₁)

= 4.5kg(0.43)(9.8m/s²)(1+4.5kg/15.0kg) = 24.65N

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F_s: static friction

a₁: acceleration of bottom (large) block

a₂: acceleration of top (small) block

µ_s: coefficient of static friction

g: acceleration due to gravity = 9.8m/s²

N₁: normal force on bottom (large) block

N₂: normal force on top (small) block