Question

A concrete block of mass 35kg is pulled along a horizontal floor with the aid of a rope inclined at an angle of 30⁰ to the horizontal. If the coefficient of friction is 0.75, Calculate the force required to move the block( g = 10 ms⁻²)
A. 401.5 N
B. 226.5 N
C. 211.5 N
D. 245. 7N​

Answer 1

Answer:

C. 211.5 N

Explanation:

We included a free body diagram representation of the concrete block. The equation of equilibrium in the x-direction:

$\Sigma F_{x} = P\cdot \cos \alpha -\mu_{k}\cdot N = m\cdot a$ (1)

$\Sigma F_{y} = N + P\cdot \sin \alpha -m\cdot g = 0$ (2)

Where:

$P$ - External force exerted on the block, measured in newtons.

$N$ - Normal force from the ground to the block, measured in newtons.

$\alpha$ - Direction of the external force, measured in sexagesimal degrees.

$\mu_{k}$ - Kinetic coefficient of friction, dimensionless.

$m$ - Mass of the concrete block, measured in kilograms.

$g$ - Gravitational acceleration, measured in meters per square second.

$a$ - Net acceleration of the concrete block, measured in meters per square second.

The minimum force to move block occurs when net force is zero, from (2) we get that the normal force is:

$N = m\cdot g -P\cdot \sin \alpha$

And by applying this formula in (1), we get that the external force is:

$P\cdot \cos \alpha - \mu_{k}\cdot (m\cdot g -P\cdot \sin \alpha) = m\cdot a$

$P\cdot \cos \alpha -\mu_{k}\cdot m\cdot g +\mu_{k}\cdot P\cdot \sin \alpha = 0$

$P\cdot (\cos \alpha +\mu_{k}\cdot \sin \alpha)=\mu_{k}\cdot m\cdot g$

$P = \frac{\mu_{k}\cdot m\cdot g}{\cos \alpha +\mu_{k}\cdot \sin \alpha}$

If we know that $\mu_{k} = 0.75$, $m = 35\,kg$, $g = 10\,\frac{m}{s^{2}}$ and $\alpha = 30^{\circ}$. then the external force is:

$P = \frac{(0.75)\cdot (35\,kg)\cdot \left(10\,\frac{m}{s} \right)}{\cos 30^{\circ}+0.75\cdot \sin 30^{\circ}}$

$P \approx 211.518\,N$

The correct answer is C.