Question

You push downward on a trunk at an angle 25° below the horizontal with a force of 750 n. if the trunk is on a flat surface and the coefficient of static friction between the surface and the trunk is 0.61, what is the most massive trunk you will be able to move?

Answer 1

N = normal force on the trunk

m = mass of the trunk

W = weight of the trunk = mg

Using equilibrium of force in vertical direction

N = W + 750 Sin25

N = mg + 750 Sin25                                       eq-1

μ = Coefficient of static friction = 0.61

f = static frictional force

static frictional force is given as

f = μN

using eq-1

f = (0.61) (mg + 750 Sin25)                           eq-2

Along the horizontal direction , for the trunk to move, force equation must be

750 Cos25 = f

using eq-2

750 Cos25 = (0.61) (mg + 750 Sin25)

750 Cos25 = (0.61) (m (9.8) + 750 Sin25)

m = 81.4 kg

Answer 2

The most massive trunk is about 81 kg

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Further explanation

Acceleration is rate of change of velocity.

$\large {\boxed {a = \frac{v - u}{t} } }$

$\large {\boxed {d = \frac{v + u}{2}~t } }$

a = acceleration (m / s²)v = final velocity (m / s)

u = initial velocity (m / s)

t = time taken (s)

d = distance (m)

Let us now tackle the problem!

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Given:

direction of force = θ = 25°

magnitude of force = F = 750 N

coefficient of station friction = μ = 0.61

Asked:

mass of trunk = m = ?

Solution:

We will solve this problem by using Newton's Law of Motion as follows:

$\Sigma F_y = 0$

$N - F \sin \theta - w = 0$

$\boxed{N = F \sin \theta + w}$ → Equation 1

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$\Sigma F_x = 0$

$F \cos \theta - f = 0$

$F \cos \theta - \mu N = 0$

$F \cos \theta - \mu ( F \sin \theta + w ) = 0$ ← Equation 1

$F \cos \theta - \mu F \sin \theta - \mu w = 0$

$F \cos \theta - \mu F \sin \theta = \mu m g$

$m = (F \cos \theta - \mu F \sin \theta) \div (\mu g)$

$m = (750 \times \cos 25^o - 0.61 \times 750 \times \sin 25^o) \div (0.61 \times 9.8)$

$\boxed{m \approx 81 \texttt{ kg}}$

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Answer details

Grade: High School

Subject: Physics

Chapter: Dynamics