What is the Magnitude of the Force on a Block on a Horizontal Surface?

Thread starter nmegabyte
Start date Apr 18, 2010
Tags

Force Magnitude

AI Thread Summary

The block, weighing 3.0 N, experiences a 0.9 N upward force from a vertical string. To find the force exerted by the block on the horizontal surface, the net downward force must be calculated. The force on the surface equals the weight of the block minus the upward force, resulting in a force of 2.1 N. Participants emphasize the importance of showing calculations to receive assistance. Understanding the balance of forces is crucial in solving the problem accurately.

Apr 18, 2010

nmegabyte

A block with a weight of 3.0 N is at rest on a horizontal surface. A 0.9 N upward force is applied to the block by means of attached vertical string.
(a) What is the magnitude of the force of the block on the horizontal surface?

Physics news on Phys.org

Apr 18, 2010

NeoDevin

You must show an attempt before we can help you.

Thread 'I've come across another fluid pressure problem I don't understand'

Neglect gravity other than that of water; neglect friction. F1=ρgsh=1000*10*1*(1+4)=50000 N; F1=ρgsh=1000*10*0.1*4=4000 N; F3=50000-4000=46000 N?

View full post »

Thread 'Struggling to understand the concept of this question (ice cube in water optics question)'

TL;DR Summary: I came across this question from a Sri Lankan A-level textbook. Question - An ice cube with a length of 10 cm is immersed in water at 0 °C. An observer observes the ice cube from the water, and it seems to be 7.75 cm long. If the refractive index of water is 4/3, find the height of the ice cube immersed in the water. I could not understand how the apparent height of the ice cube in the water depends on the height of the ice cube immersed in the water. Does anyone have an...

View full post »

Thread 'Variable mass system : water sprayed into a moving container'

Starting with the mass considerations #m(t)# is mass of water #M_{c}# mass of container and #M(t)# mass of total system $$M(t) = M_{C} + m(t)$$ $$\Rightarrow \frac{dM(t)}{dt} = \frac{dm(t)}{dt}$$ $$P_i = Mv + u \, dm$$ $$P_f = (M + dm)(v + dv)$$ $$\Delta P = M \, dv + (v - u) \, dm$$ $$F = \frac{dP}{dt} = M \frac{dv}{dt} + (v - u) \frac{dm}{dt}$$ $$F = u \frac{dm}{dt} = \rho A u^2$$ from conservation of momentum , the cannon recoils with the same force which it applies. $$\quad \frac{dm}{dt}...

View full post »