an 85kg concrete block rests on a flat surface with coefficient of static friction us=0.42. find the maximum force that can pull on the block with, angle of 38 degrees above the horizontal without moving the block?

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pulling at an angle reduces the frictional force (by reducing the normal force)

so the frictional force is
... .42 * {(85 * g) - [F sin(38º)]}

the block is in equilibrium when the frictional force equals the horizontal component of the pulling force
... F cos(38º)

set the two expressions equal, and solve for F

To find the maximum force that can be applied to the block without moving it, we need to consider the forces acting on the block and their relationship to friction.

The maximum force that can be applied to the block without moving it is equal to the product of the coefficient of static friction (μs) and the normal force (F_N) acting on the block:

Maximum Force (F_max) = μs * F_N

To find the normal force, we need to consider the weight of the block and the angle of the force being applied. Assuming that the force is applied perpendicular to the surface, we can use trigonometry to determine the normal force.

The weight of the block (W) is given by the formula:

W = m * g

where m is the mass of the block (85 kg) and g is the acceleration due to gravity (approximately 9.8 m/s^2).

W = 85 kg * 9.8 m/s^2
= 833 N

To find the normal force (F_N), we need to resolve the weight force into its components. The normal force is the component of the weight force acting perpendicular to the surface, which can be found using trigonometry:

F_N = W * cos(θ)

where θ is the angle of the force above the horizontal (38 degrees).

F_N = 833 N * cos(38°)
≈ 666.97 N

Finally, we can find the maximum force that can be applied to the block without moving it:

F_max = μs * F_N
= 0.42 * 666.97 N
≈ 280.19 N

Therefore, the maximum force that can be applied to the block without moving it, at an angle of 38 degrees above the horizontal, is approximately 280.19 N.