Question

Two blocks connected by a string are pulled across a horizontal surface by a force applied to one of the blocks, as shown below. The coefficient of kinetic friction between the blocks and the surface is 0.25. If each block has an acceleration of 3.0 m/s² to the right, what is the magnitude F of the applied force (in N)?

M1=1.2 kg
M2=3.2 kg
Angle that F makes with the horizontal=69⁰

Answer 1

77 N

Step-by-step explanation:

Draw free body diagrams for each block.

The smaller block has four forces acting on it:

Weight force m₁g pulling down,

Normal force N₁ pushing up,

Friction force μN₁ pushing to the left,

and tension force T pulling to the right.

Sum of forces in the y direction:

∑F = ma

N₁ − m₁g = 0

N₁ = m₁g

N₁ = (1.2 kg) (9.8 m/s²)

N₁ = 11.76 N

Sum of forces in the x direction:

∑F = ma

T − μN₁ = m₁a

T = m₁a + μN₁

T = (1.2 kg) (3.0 m/s²) + (0.25) (11.76 N)

T = 6.54 N

The larger block has five forces acting on it:

Weight force m₂g pulling down,

Normal force N₂ pushing up,

Friction force μN₂ pushing to the left,

Tension force T pulling to the left,

and applied force F pulling 69° above the horizontal.

Sum of forces in the y direction:

∑F = ma

N₂ + F sin 69° − m₂g = 0

N₂ + F sin 69° = m₂g

N₂ + F sin 69° = (3.2 kg) (9.8 m/s²)

N₂ + F sin 69° = 31.36 N

Sum of forces in the x direction:

∑F = ma

F cos 69° − T − μN₂ = m₂a

F cos 69° − μN₂ = m₂a + T

F cos 69° − 0.25 N₂ = (3.2 kg) (9.8 m/s²) + 6.54 N

F cos 69° − 0.25 N₂ = 37.9 N

We have two equations and two variables. Solve for N₂ in the first equation and substitute into the second.

N₂ = 31.36 − F sin 69°

F cos 69° − 0.25 (31.36 − F sin 69°) = 37.9

F cos 69° − 7.84 + 0.25 F sin 69° = 37.9

F (cos 69° +0.25 sin 69°) = 45.74

0.592 F = 45.74

F = 77.3 N

Rounded to two significant figures, the magnitude of the force is 77 N.